User manual | Chapter 13 Quadratic Relations and Functions

CHAPTER
13
CHAPTER
TABLE OF CONTENTS
13-1 Solving Quadratic Equations
13-2 The Graph of a Quadratic
Function
13-3 Finding Roots from a Graph
13-4 Graphic Solution of a
Quadratic-Linear System
13-5 Algebraic Solution of a
Quadratic-Linear System
Chapter Summary
Vocabulary
Review Exercises
Cumulative Review
502
QUADRATIC
RELATIONS
AND
FUNCTIONS
When a baseball is hit, its path is not a straight line.
The baseball rises to a maximum height and then falls,
following a curved path throughout its flight.The maximum height to which it rises is determined by the
force with which the ball was hit and the angle at
which it was hit.The height of the ball at any time can
be found by using an equation, as can the maximum
height to which the ball rises and the distance between
the batter and the point where the ball hits the
ground.
In this chapter we will study the quadratic equation
that models the path of a baseball as well as functions
and relations that are not linear.
Solving Quadratic Equations
503
13-1 SOLVING QUADRATIC EQUATIONS
The equation x2 3x 10 0 is a polynomial equation in one variable. This
equation is of degree two, or second degree, because the greatest exponent of
the variable x is 2. The equation is in standard form because all terms are collected in descending order of exponents in one side, and the other side is 0.
A polynomial equation of degree two is also called a quadratic equation.
The standard form of a quadratic equation in one variable is
ax2 bx c 0
where a, b, and c are real numbers and a 0.
To write an equation such as x(x 4) 5 in standard form, rewrite the left
side without parentheses and add 5 to both sides to make the right side 0.
x(x 4) 5
x2 4x 5
x2 4x 5 0
Solving a Quadratic Equation by Factoring
When 0 is the product of two or more factors, at least one of the factors must be
equal to 0. This is illustrated by the following examples:
500
(2) 0 0
1
2
070
0 (3) 0
000
3050
In general:
When a and b are real numbers, ab 0 if and only if a 0 or b 0.
This principle is used to solve quadratic equations. For example, to solve the
quadratic equation x2 3x 2 0, we can write the left side as (x 2)(x 1).
The factors (x 2 2) and (x 1) represent real numbers whose product is 0. The
equation will be true if the first factor is 0, that is, if (x 2) 0 or if the second
factor is 0, that is, if (x 1) 0.
x2 3x 2 0
(x 2)(x 1) 0
x2 0
2 2
x
2
x1 0
1 1
x
1
A check will show that both 2 and 1 are values of x for which the equation
is true.
504
Quadratic Relations and Functions
Check for x 2:
Check for x 1:
x2 3x 2 0
x2 3x 2 0
?
(1)2 3(1) 2 5 0
?
?
1 3 2 50
(2)2 3(2) 2 5 0
?
4 6 2 50
00✔
00✔
Since both 2 and 1 satisfy the equation x 3x 2 0, the solution set of
this equation is {2, 1}. The roots of the equation, that is, the values of the variable that make the equation true, are 2 and 1.
Note that the factors of the trinomial x2 3x 2 are (x 2) and (x 1).
If the trinomial x2 3x 2 is set equal to zero, then an equation is formed and
this equation has a solution set, {2, 1}.
2
Procedure
To solve a quadratic equation by factoring:
1. If necessary, transform the equation into standard form.
2. Factor the quadratic expression.
3. Set each factor containing the variable equal to 0.
4. Solve each of the resulting equations.
5. Check by substituting each value of the variable in the original equation.
The real number k is a root of ax2 bx c 0 if and only if (x k) is a
factor of ax2 bx c.
EXAMPLE 1
Solve and check: x2 7x 10
Solution
How to Proceed
x2 7x 10
(1) Write the equation in standard form:
x2 7x 10 0
(2) Factor the quadratic expression:
(3) Let each factor equal 0:
(4) Solve each equation:
(x 2)(x 5) 0
x20
x50
x2
x5
Solving Quadratic Equations
505
(5) Check both values in the original equation:
Check for x 2:
Check for x 5:
x2 7x 10
x2 7x 10
?
(5)2 7(5) 5 210
?
25 35 5 210
(2)2 7(2) 5 210
4 14 5 210
10 10 ✔
?
?
10 10 ✔
Answer x 2 or x 5; the solution set is {2, 5}.
EXAMPLE 2
List the members of the solution set of 2x2 3x.
Solution
How to Proceed
2x2 3x
(1) Write the equation in standard form:
2x2 3x 0
(2) Factor the quadratic expression:
(3) Let each factor equal 0:
x(2x 3) 0
x0
2x 3 0
2x 3
(4) Solve each equation:
x 5 32
(5) Check both values in the original equation.
Check for x 0:
Check for x 32:
2x2 3x
2x2 3x
?
2(0)2 5 3(0)
00✔
2 A 32 B 5 3 A 32 B
2 ?
2 A 94 B 5 3 A 32 B
?
9
2
5 92 ✔
Answer: The solution set is U0, 32 V .
Note: We never divide both sides of an equation by an expression containing a
variable. If we had divided 2x2 = 3x by x, we would have obtained the equation
2x 3, whose solution is 32 but would have lost the solution x 0.
506
Quadratic Relations and Functions
EXAMPLE 3
Find the solution set of the equation x(x 8) 2x 25.
Solution
How to Proceed
x(x 8) 2x 25
x2 8x 2x 25
(1) Use the distributive property on the
left side of the equation:
x2 10x 25 0
(2) Write the equation in standard form:
(x 5)(x 5) 0
(3) Factor the quadratic expression:
(4) Let each factor equal 0:
x50
x50
x5
x5
(5) Solve each equation:
x(x 8) 2x 25
(6) Check the value in the original equation:
?
5(5 8) 5 2(5) 25
?
15 5 10 2 25
15 15 ✔
Answer x 5; the solution set is {5}.
Every quadratic equation has two roots, but as Example 3 shows, the two
roots are sometimes the same number. Such a root, called a double root, is written only once in the solution set.
EXAMPLE 4
The height h of a ball thrown into the air with an initial vertical velocity of 24
feet per second from a height of 6 feet above the ground is given by the equation h l6t2 24t 6 where t is the time, in seconds, that the ball has been in
the air. After how many seconds is the ball at a height of 14 feet?
Solution (1) In the equation, let h 14:
h l6t 2 24t 6
14 16t 2 24t 6
(2) Write the equation in standard form:
0 16t 2 24t 8
(3) Factor the quadratic expression:
0 8(2t 2 3t 1)
0 8(2t 1)(t 1)
(4) Solve for t:
2t 1 0
t10
2t 1
t1
t5
1
2
Answer The ball is at a height of 14 feet after 12 second as it rises and after 1 second
as it falls.
Solving Quadratic Equations
507
EXAMPLE 5
The area of a circle is equal to 3 times its circumference. What is the radius of
the circle?
How to Proceed
Solution
(1)
(2)
(3)
(4)
(5)
Write an equation from the given information:
Set the equation in standard form:
Factor the quadratic expression:
Solve for r:
Reject the zero value. Use the positive
value to write the answer.
pr2 3(2pr)
pr 6pr 0
pr(r 6) 0
pr 0
r60
r0
r6
2
Answer The radius of the circle is 6 units.
EXERCISES
Writing About Mathematics
1. Can the equation x2 9 be solved by factoring? Explain your answer.
2. In Example 4, the trinomial was written as the product of three factors. Only two of these
factors were set equal to 0. Explain why the third factor was not used to find a solution of
the equation.
Developing Skills
In 3–38, solve each equation and check.
3. x2 3x 2 0
4. z2 5z 4 0
5. x2 8x 16 0
6. r2 12r 35 0
7. c2 6c 5 0
8. m2 10m 9 0
9. x2 2x 1 0
10. y2 11y 24 0
11. x2 4x 5 0
12. x2 x 6 0
13. x2 2x 15 0
14. r2 r 72 0
15. x2 x 12 0
16. x2 49 0
17. z2 4 0
18. m2 64 0
19. 3x2 12 0
20. d2 2d 0
21. s2 s 0
22. x2 3x 0
23. z2 8z 0
24. x2 x 6
25. y2 3y 28
26. c2 8c 15
27. r2 4
28. x2 121
29. y2 6y
30. s2 4s
31. y2 8y 20
32. x2 9x 20
33. 30 x x2
34. x2 3x 4 50
35. 2x2 7 5 5x
36. x(x 2) 35
37. y(y 3) 4
38. x(x 3) 40
508
Quadratic Relations and Functions
In 39–44, solve each equation and check.
2
12
39. x 1
2 5 x
40.
14
42. x21
5 x4
43.
y13
3
24x
x23
5 y6
41. x3 5 12
x
1
5 x2
2
22
x21
44. 2x
x13 5 x22
Applying Skills
45. The height h of a ball thrown into the air with an initial vertical velocity of 48 feet per second from a height of 5 feet above the ground is given by the equation
h 16t2 48t 5
where t is the time in seconds that the ball has been in the air. After how many seconds is
the ball at a height of 37 feet?
46. A batter hit a baseball at a height of 4 feet with a force that gave the ball an initial vertical
velocity of 64 feet per second. The equation
h 16t2 64t 4
gives the height h of the baseball t seconds after the ball was hit. If the ball was caught at a
height of 4 feet, how long after the batter hit the ball was the ball caught?
47. The length of a rectangle is 12 feet more than twice the width. The area of the rectangle is
320 square feet.
a. Write an equation that can be used to find the length and width of the rectangle.
b. What are the dimensions of the rectangle?
48. A small park is enclosed by four streets, two of which are parallel. The park is in the shape
of a trapezoid. The perpendicular distance between the parallel streets is the height of the
trapezoid. The portions of the parallel streets that border the park are the bases of the
trapezoid. The height of the trapezoid is equal to the length of one of the bases and 20 feet
longer than the other base. The area of the park is 9,000 square feet.
a. Write an equation that can be used to find the height of the trapezoid.
b. What is the perpendicular distance between the two parallel streets?
49. At a kennel, each dog run is a rectangle whose length is 4 feet more than twice the width.
Each run encloses 240 square feet. What are the dimensions of the runs?
50. One leg of a right triangle is 14 centimeters longer than the other leg. The length of the
hypotenuse is 26 centimeters. What are the lengths of the legs?
13-2 THE GRAPH OF A QUADRATIC FUNCTION
A batter hits a baseball at a height of 3 feet off the ground, with an initial vertical velocity of 72 feet per second. The height, y, of the baseball can be found
using the equation y 5 216x2 1 72x 1 3 when x represents the number of
seconds from the time the ball was hit. The graph of the equation––and, inci-
The Graph of a Quadratic Function
509
dentally, the actual path of the ball––is a curve called a parabola. The special
properties of parabolas are discussed in this section.
An equation of the form y ax2 bx c (a 0) is called a second-degree
polynomial function or a quadratic function. It is a function because for every
ordered pair in its solution set, each different value of x is paired with one and
only one value of y. The graph of any quadratic function is a parabola.
Because the graph of a quadratic function is nonlinear, a larger number of
points are needed to draw the graph than are needed to draw the graph of a
linear function. The graphs of two equations of the form y ax2 bx c, one
that has a positive coefficient of x2 (a 0) and the other a negative coefficient
of x2 (a 0), are shown below.
CASE 1
The graph of ax2 bx c where a 0.
Graph the quadratic function y x2 4x 1 for integral values of x from
1 to 5 inclusive:
(1) Make a table using integral values of x from 1 to 5.
(2) Plot the points associated with each ordered pair (x, y).
(3) Draw a smooth curve through the points.
y
1
y
–4x+
x2 4x 1
1
141
6
0
001
1
1
141
2
2
481
3
1
3
9 12 1
2
4
16 16 1
1
–1 O
–1
5
25 20 1
6
y = x2
x
1
x
The values of x that were chosen to draw this graph are not a random set of
numbers. These numbers were chosen to produce the pattern of y-values shown
in the table. Notice that as x increases from 1 to 2, y decreases from 6 to 3.
Then the graph reverses and as x continues to increase from 2 to 5, y increases
from 3 to 6. The smallest value of y occurs at the point (2, 3). The point is
called the minimum because its y-value, 3, is the smallest value of y for the
equation. The minimum point is also called the turning point or vertex of the
parabola.
510
Quadratic Relations and Functions
The graph is symmetric with respect to the vertical line, called the axis of
symmetry of the parabola. The axis of symmetry of the parabola is determined
by the formula
x 5 2b
2a
where a and b are the coefficients x2 and x, respectively, from the standard form
of the quadratic equation.
For the function y x2 4x 1, the equation of the vertical line of sym2(24)
metry is x 5 2(1) or x 2. Every point on the parabola to the left of x 2
matches a point on the parabola to the right of x 2, and vice versa.
This example illustrates the following properties of the graph of the quadratic equation y x2 4x 1:
1. The graph of the equation is a parabola.
2. The parabola is symmetric with respect to the vertical line x 2.
3. The parabola opens upward and has a minimum point at (2, 3).
4. The equation defines a function. For every x-value there is one and only
one y-value.
5. The constant term, 1, is the y-intercept. The y-intercept is the value of y
when x is 0.
CASE 2
The graph of ax2 bx c where a 0.
Graph the quadratic function y x2 2x 5 using integral values of x
from 4 to 2 inclusive:
(1) Make a table using integral values of x from 4 to 2.
(2) Plot the points associated with each ordered pair (x, y).
(3) Draw a smooth curve through the points.
y
4
16 8 5
3
3
9 6 5
2
2
4 4 5
5
1
1 2 5
6
0
005
5
1
1 2 5
2
2
4 4 5
3
1
–1
–1
O
1
5
y
x+
x2 – 2
x2 2x 5
y=–
x
x
Again, the values of x that were chosen to produce the pattern of y-values
shown in the chart. Notice that as x increases from 4 to 1, y increases from
The Graph of a Quadratic Function
511
3 to 6. Then the graph reverses, and as x continues to increase from 1 to 2,
y decreases from 6 to 3. The largest value of y occurs at the point (1, 6). This
point is called the maximum because its y-value, 6, is the largest value of y for
the equation. In this case, the maximum point is the turning point or vertex of
the parabola.
The graph is symmetric with respect to the vertical line whose equation is
x 1. As shown in Case 1, this value of x is again given by the formula
x 5 2b
2a
where a and b are the coefficients of x2 and x, respectively. For the function
2(22)
y x2 2x 5, the equation of the axis of symmetry is x 2(21) or x 1.
This example illustrates the following properties of the graph of the quadratic equation y 5 2x2 2 2x 1 5:
1. The graph of the equation is a parabola.
2. The parabola is symmetric with respect to the vertical line x 1.
3. The parabola opens downward and has a maximum point at (1, 6)
4. The equation defines a function.
5. The constant term, 5, is the y-intercept.
When the equation of a parabola is written in the form y ax2 bx c 0,
the equation of the axis of symmetry is x 2b
2a and the x-coordinate of the turn2b
ing point is 2a . This can be used to find a convenient set of values of x to be used
when drawing a parabola. Use 2b
2a as a middle value of x with three values that
are smaller and three that are larger.
KEEP IN MIND
1. The graph of y ax2 bx c, with a 0, is a parabola.
2. The axis of symmetry of the parabola is a vertical line whose equation is
x 2b
2a .
3. A parabola has a turning point on the axis of symmetry. The x-coordinate
of the turning point is 2b
2a . The y-coordinate of the turning point is found
by substituting 2b
into
the equation of the parabola. The turning point is
2a
the vertex of the parabola.
4. If a is positive, the parabola opens upward and the turning point is a minimum. The minimum value of y for the parabola is the y-coordinate of the
turning point.
5. If a is negative, the parabola opens downward and the turning point is a
maximum. The maximum value of y for the parabola is the y-coordinate of
the turning point.
512
Quadratic Relations and Functions
EXAMPLE 1
a. Write the equation of the axis of symmetry of y x2 3.
b. Graph the function.
c. Does the function have a maximum or a minimum?
d. What is the maximum or minimum value of the function?
e. Write the coordinates of the vertex.
Solution a. In this equation, a 1. Since there is no x term in the equation, the equation can be written as y x2 0x 3 with b 0. The equation of the axis
of symmetry is
2(0)
x 2b
2a 5 2(1) 0 or x 0.
b. (1) Since the vertex of the parabola is on the axis of symmetry, the x-coordinate of the vertex is 0. Use three values of x that are less than 0 and
three values of x that are greater than 0. Make a table using integral values of x from 3 to 3.
(2) Plot the points associated with each ordered pair (x, y).
(3) Draw a smooth curve through the points to draw a parabola.
y
x2 3
y
–3
(3)2 3
6
–2
(2) 3
1
–1
(1) 3
2
0
(0) 3
3
1
1
(1) 3
2
2
(2) 3
1
–1
–1
3
(3) 3
6
2
2
2
2
2
y = x2
2
–3
x
O
1
x
c. Since a 1 0, the function has a minimum.
d. The minimum value of the function is the y-coordinate of the vertex, 3,
which can be read from the table of values.
e. Since the vertex is the turning point of this parabola, the coordinates of the
vertex are (0, 3).
The Graph of a Quadratic Function
513
Calculator a. Determine the equation of the axis of symmetry as before.
Solution b. Enter the equation in the Y list of functions and graph the function. Clear
any equations already in the list.
ENTER:
DISPLAY:
Y
X,T,,n
x2
3 ZOOM
6
Plot1 Plot2 Plot3
\Y 1 = X 2 –3
\Y 2 =
\Y 3 =
\Y =
\Y 45 =
\Y 6 =
\Y 7 =
c. The graph shows that the function has a minimum.
d. Since the minimum of the function occurs at the vertex, use value
CALC
1 R from the CALC menu to evaluate the
Q 2nd
function at x 0, the x-coordinate of the vertex:
ENTER:
DISPLAY:
2nd
CALC
1
0 ENTER
Calculate
1: value
2: zero
3: minimum
4: minimum
5: intersect
6: dy/dx
7: f [x] dx
Y1 = X2–3
*
X= 0
Y = –3
The calculator displays the minimum value, 3.
e. The coordinates of the vertex are (0, 3).
Answers a. The axis of symmetry is the y-axis. The equation is x 0.
b. Graph
c. The function has a minimum.
d. The minimum value is 3.
e. The vertex is (0, 3).
514
Quadratic Relations and Functions
When the coordinates of the turning point are rational numbers, we can use
CALC
CALC
4 R from the
3 R or maximum Q 2nd
minimum Q 2nd
CALC menu to find the vertex. In Example 1, since the turning point is a minimum, use minimum:
ENTER:
2nd
CALC
3
When the calculator asks “LeftBound?” move the cursor to any point to the left
of the vertex using the left or right arrow keys, and then press enter. When the
calculator asks “RightBound?” move the cursor to the right of the vertex, and
then press enter. When the calculator asks “Guess?” move the cursor near the
vertex and then press enter. The calculator displays the coordinates of the vertex at the bottom of the screen.
Y1=X^2–3
*
*
Y1=X^2–3
LeftBound?
RightBound?
X = -.8510638
Y = –2.27569
X = 1.0638298
Y = –1.868266
*
*
Y1=X^2–3
Minimum
Guess?
X=0
Y = -3
X=0
Y = -3
EXAMPLE 2
Sketch the graph of the function y x2 7x 10.
Solution (1) Find the equation of the axis of symmetry and the x-coordinate of the
vertex:
2(27)
x 2b
2a 2(1) 3.5
(2) Make a table of values using three integral values smaller than and three
larger than 3.5.
(3) Plot the points whose coordinates are given in the table and draw a
smooth curve through them.
The Graph of a Quadratic Function
515
y
x2 7x 10
x
y
1
1 7 10
4
2
4 14 10
0
3
9 21 10
2
3.5
12.25 24.5 10
1
2.25
4
16 28 10
2
5
25 35 10
0
6
36 42 10
4
O
–1
–1
x
1
y = x2 – 7x + 10
Note: The table can also be displayed on the calculator. First enter the equation into Y1.
ENTER:
Y
X,T,,n
x2
7 X,T,,n
10
Then enter the starting value and the interval between the x values. We will use
1 as the starting value and 0.5 as the interval in order to include 3.5, the x-value
of the vertex.
ENTER:
2nd
TBLSET 1 ENTER
.5 ENTER
Before creating the table, make sure that
“Indpnt:” and “Depend:” are set to “auto.” If
ENTER .
they are not, press ENTER
X
1
1.5
2
2.5
3
3.5
4
TABLE to create the
Finally, press 2nd
table. Scroll up and down to view the values
of x and y.
Y1
4
1.75
0
-1.25
-2
-2.25
-2
X=1
Calculator Enter the equation in the Y= menu and
Solution sketch the graph of the function in the standard window.
ENTER:
DISPLAY:
Y
X,T,,n
x2
7 X,T,,n
10 ZOOM
6
516
Quadratic Relations and Functions
EXAMPLE 3
The perimeter of a rectangle is 12. Let x represent the measure of one side of
the rectangle and y represent the area.
a. Write an equation for the area of the rectangle in terms of x.
b. Draw the graph of the equation written in a.
c. What is the maximum area of the rectangle?
P 2l 2w
Solution a. Let x be the measure of the length of the
rectangle. Use the formula for perimeter to
express the measure of the width in terms of x:
12 2x 2w
6xw
w6x
A lw
Write the formula for area in terms of x and y:
y x(6 x)
y 6x x2
y x2 6x
26
b. The equation of the axis of symmetry is x 2(21)
or x 3. Make a table of
values using values of x on each side of 3.
y
y
0
0 0
0
1
1 6
5
2
4 12
8
3
9 18
9
4
16 24
8
5
25 30
5
6
36 36
0
6x
x2 +
x2 6x
y=–
x
1 O
–1
–1
1
x
c. The maximum value of the area, y, is 9.
Note: The graph shows all possible values of x and y. Since both the measure
of a side of the rectangle, x, and the area of the rectangle, y, must be positive,
0 x 6 and 0 y 9. Since (2, 8) is a point on the graph, one possible rectangle has dimensions 2 by (6 2) or 2 by 4 and an area of 8. The rectangle
with maximum area, 9, has dimensions 3 by (6 3) or 3 by 3, a square.
517
The Graph of a Quadratic Function
Translating, Reflecting, and Scaling Graphs of
Quadratic Functions
Just as linear and absolute value functions can be translated, reflected, or scaled,
graphs of quadratic functions can also be manipulated by working with the
graph of the quadratic function y x2.
For instance, the graph of y 5 x2 1 2.5 is the graph of y 5 x2 shifted 2.5
units up. The graph of y x2 is the graph of y x2 reflected in the x-axis. The
graph of y 5 3x2 is the graph of y 5 x2 stretched vertically by a factor of 3,
while the graph of y 5 13x2 is the graph of y 5 x2 compressed vertically by a factor of 13.
y
1
1
–2 –1
x2
1
y=
3
y = 3 x2
y = x2
+ 2.5
1
y = x2
–1
y
y = x2
y = x2
y
x
2
y = – x2
x
2
–1
1
x
Translation Rules for Quadratic Functions
If c is positive:
The graph of y 5 x 2 1 c is the graph of y 5 x 2 shifted c units up.
The graph of y 5 x 2 2 c is the graph of y 5 x 2 shifted c units down.
The graph of y 5 (x 1 c) 2 is the graph of y 5 x 2 shifted c units to the left.
The graph of y 5 (x 2 c) 2 is the graph of y 5 x 2 shifted c units to the right.
Reflection Rule for Quadratic Functions
The graph of y 5 2x 2 is the graph of y 5 x 2 reflected in the x-axis.
Scaling Rules for Quadratic Functions
When c 1, the graph of y 5 cx 2 is the graph of y 5 x 2 stretched vertically
by a factor of c.
When 0 c 1, the graph of y 5 cx 2 is the graph of y 5 x 2 compressed
vertically by a factor of c.
518
Quadratic Relations and Functions
EXAMPLE 4
In a–e, write an equation for the resulting function if the graph of y 5 x2 is:
a. shifted 5 units down and 1.5 units to the left
b. stretched vertically by a factor of 4 and shifted 2 units down
c. compressed vertically by a factor of 16 and reflected in the x-axis
d. reflected in the x-axis, shifted 2 units up, and shifted 2 units to the right
Solution a. y 5 (x 1 1.5) 2 2 5 Answer
y 5 4x2
b. First, stretch vertically by a factor of 4:
y 5 4x2 2 2 Answer
Then, translate the resulting function
2 units down:
c. First, compress vertically by a factor of 16:
y 16x2
y 216x2 Answer
Then, reflect in the x-axis:
y 5 2x2
d. First, reflect in the x-axis:
Then, translate the resulting function
2 units up:
y 5 2x2 1 2
Finally, translate the resulting function
2 units to the right:
y 5 2(x 2 2) 2 1 2 Answer
EXERCISES
Writing About Mathematics
y
1 2
4x
1. Penny drew the graph of h(x) 5
2 2x from
x 4 to x 4. Her graph is shown to the
right. Explain why Penny’s graph does not look
like a parabola.
1
O
1
x
The Graph of a Quadratic Function
519
2. What values of x would you choose to draw the graph of h(x) 5 14x2 2 2x so that points on
both sides of the turning point would be shown on the graph? Explain your answer.
Developing Skills
In 3–14: a. Graph each quadratic function on graph paper using the integral values for x indicated
in parentheses to prepare the necessary table of values. b. Write the equation of the axis of symmetry of the graph. c. Write the coordinates of the turning point of the graph.
3. y x2 (3 x 3)
4. y x2 (3 x 3)
5. y x2 1 (3 x 3)
6. y x2 1 (3 x 3)
7. y x2 4 (3 x 3)
8. y x2 2x (2 x 4)
9. y x2 2x (2 x 4)
10. y x2 6x 8 (0 x 6)
11. y x2 4x 3 (1 x 5)
12. y x2 2x 1 (2 x 4)
13. y x2 2x 3 (4 x 2)
14. y x2 4x 3 (1 x 5)
In 15–20: a. Write the equation of the axis of symmetry of the graph of the function. b. Find the
coordinates of the vertex. c. Draw the graph on graph paper or on a calculator, showing at least
three points with integral coefficients on each side of the vertex.
15. y x2 6x 1
16. y x2 2x 8
17. y x2 8x 12
18. y x2 4x 3
19. y x2 3x 7
20. y x2 x 5
21. Write an equation for the resulting function if the graph of y x2 is:
a. reflected in the x-axis and shifted 3 units left.
b. compressed vertically by a factor of 27 and shifted 9 units up.
c. reflected in the x-axis, stretched vertically by a factor of 6, shifted 1 unit down, and
shifted 4 units to the right.
In 22–25, each graph is a translation and/or a reflection of the graph of y x2. For each graph,
a. determine the vertex and the axis of symmetry, and b. write the equation of each graph.
22.
23.
y
x
y
x
520
Quadratic Relations and Functions
24.
25.
y
y
x
x
26. Of the graphs below, which is the graph of a quadratic function and the graph of an absolute
value function?
(1)
y
(3)
y
x
x
(2)
(4)
y
x
y
x
Applying Skills
27. The length of a rectangle is 4 more than its width.
a. If x represents the width of the rectangle, represent the length of the rectangle in terms
of x.
b. If y represents the area of the rectangle, write an equation for y in terms of x.
c. Draw the graph of the equation that you wrote in part b.
d. Do all of the points on the graph that you drew represent pairs of values for the width
and area of the rectangle? Explain your answer.
The Graph of a Quadratic Function
521
28. The height of a triangle is 6 less than twice the length of the base.
a. If x represents the length of the base of the triangle, represent the height in terms of x.
b. If y represents the area of the triangle, write an equation for y in terms of x.
c. Draw the graph of the equation that you wrote in part b.
d. Do all of the points on the graph that you drew represent pairs of values for the length
of the base and area of the triangle? Explain your answer.
29. The perimeter of a rectangle is 20 centimeters. Let x represent the measure of one side of
the rectangle and y represent the area of the rectangle.
a. Use the formula for perimeter to express the measure of a second side of the
rectangle.
b. Write an equation for the area of the rectangle in terms of x.
c. Draw the graph of the equation written in b.
d. What are the dimensions of the rectangle with maximum area?
e. What is the maximum area of the rectangle?
f. List four other possible dimensions and areas for the rectangle.
30. A batter hit a baseball at a height 3 feet off the ground, with an initial vertical velocity of
64 feet per second. Let x represent the time in seconds, and y represent the height of the
baseball. The height of the ball can be determined over a limited period of time by using the
equation y 16x2 64x 3.
a. Make a table using integral values of x from 0 to 4 to find values of y.
b. Graph the equation. Let one horizontal unit 14 second, and one vertical unit 10
feet. (See suggested coordinate grid below.)
h
100
80
60
40
20
0
0
1
2
3
4
5
6
t
c. If the ball was caught after 4 seconds, what was its height when it was caught?
d. From the table and the graph, determine:
(1) the maximum height reached by the baseball;
(2) the time required for the ball to reach this height.
522
Quadratic Relations and Functions
13-3 FINDING ROOTS FROM A GRAPH
In Section 1, you learned to find the solution of an equation of the form
ax2 bx c 0 by factoring. In Section 2, you learned to draw the graph of a
function of the form y 5 ax2 1 bx 1 c. How are these similar expressions
related? To answer this question, we will consider three possible cases.
A quadratic equation can have two distinct real roots.
+ 10
– 7x
1
–1
–1
O
x
1
A quadratic equation can have only one distinct real root.
y
+9
The two roots of the equation
x2 6x 9 0 are equal.There is only
one number, 3, that makes the equation true. The function y x2 6x 9
has infinitely many pairs of numbers
that make the equation true. The
graph of this function shown at the
right intersects the x-axis in only one
point, (3, 0). Since the y-coordinate
of this point is 0, the x-coordinate of
this point is the root of the equation
x2 6x 9 0.
– 6x
CASE 2
y
y = x2
The equation x2 7x 10 0
has exactly two roots or solutions,
5 and 2, that make the equation true.
The function, y x2 7x 10 has
infinitely many pairs of numbers that
make the equation true. The graph of
this function shown at the right intersects the x-axis in two points, (5, 0)
and (2, 0). Since the y-coordinates of
these points are 0, the x-coordinates
of these points are the roots of the
equation x2 7x 10 0.
1
–1
–1
y = x2
CASE 1
O
1
x
Recall from Section 1 that when a quadratic equation has only one distinct
root, the root is said to be a double root. In other words, when the root of a quadratic equation is a double root, the graph of the corresponding quadratic function intersects the x-axis exactly once.
523
Finding Roots from a Graph
A quadratic equation can have no real roots.
+3
y
– 2x
The equation x2 2x 3 0 has
no real roots. There is no real number
that makes the equation true. The
function, y x2 2x 3 has infinitely many pairs of numbers that
make the equation true. The graph of
this function shown at the right does
not intersect the x-axis. There is no
point on the graph whose y-coordinate is 0. Since there is no point
whose y-coordinate is 0, there are no
real numbers that are roots of the
equation x2 2x 3 0.
y = x2
CASE 3
1
–1
–1
O
1
x
The equation of the x-axis is y 0. The x-coordinates of the points at which
the graph of y ax2 bx c intersects the x-axis are the roots of the equation
ax2 bx c 0. The graph of the function y ax2 bx c can intersect the
x-axis in 0, 1, or 2 points, and the equation ax2 bx c 0 can have 0, 1, or 2
real roots.
A real number k is a root of the quadratic equation ax2 bx c 0 if and
only if the graph of y ax2 bx c intersects the x-axis at (k, 0).
EXAMPLE 1
Use the graph of y x2 5x 6 to find
the roots of x2 5x 6 0.
y
– 5x
+6
Solution The graph intersects the x-axis at (2, 0)
and (3, 0).
Answer The roots are 2 and 3. The solution set is
{2, 3}.
y = x2
The x-coordinates of these points are the
roots of x2 5x 6 0.
1
–1
–1
O
1
x
Note that in Example 1 the quadratic expression x2 5x 6 can be factored
into (x 2)(x 3), from which we can obtain the solution set 52, 36 .
524
Quadratic Relations and Functions
EXAMPLE 2
Use the graph of y x2 3x 4 to find
the linear factors of x2 1 3x 2 4.
y
Solution The graph intersects the x-axis at (4, 0)
and (1, 0).
Therefore, the roots are 4 and 1.
If x 4, then x (4) (x 4) is a
factor.
1O
–1
–1
x
If x 1, then (x 1) is a factor.
Answer The linear factors of x2 3x 4 are
(x 4) and (x 1).
y = x2 + 3x – 4
EXERCISES
Writing About Mathematics
1. The coordinates of the vertex of a parabola y x2 2x 5 are (1, 4). Does the equation
x2 2x 5 0 have real roots? Explain your answer.
2. The coordinates of the vertex of a parabola y x2 2x 3 are (1, 4). Does the equation x2 2x 3 0 have real roots? Explain your answer.
Developing Skills
In 3–10: a. Draw the graph of the parabola. b. Using the graph, find the real numbers that are elements of the solution set of the equation. c. Using the graph, factor the corresponding quadratic
expression if possible.
3. y x2 6x 5; 0 x2 6x 5
4. y x2 2x 1; 0 x2 2x 1
5. y x2 2x 3; 0 x2 2x 3
6. y x2 x 2; 0 x2 x 2
7. y x2 2x 1; 0 x2 2x 1
8. y x2 3x 2; 0 x2 3x 2
9. y x2 4x 5; 0 x2 4x 5
10. y x2 5x 6; 0 x2 5x 6
11. If the graph of a quadratic function, f(x), crosses the x-axis at x 6 and x 8, what are two
factors of f(x)?
12. If the factors of a quadratic function, h(x), are (x 6) and (x 3), what is the solution set
for the equation h(x) 0?
525
Graphic Solution of a Quadratic-Linear System
In 13–15, refer to the graph of the parabola shown below.
13. Which of the following is the equation of the parabola?
y
(1) y (x 2)(x 3)
(2) y (x 2)(x 3)
(3) y (x 2)(x 3)
(4) y (x 2)(x 3)
1
14. Set the equation of the parabola equal to 0. What are the
roots of this quadratic equation?
O
x
1
15. If the graph of the parabola is reflected in the x-axis, how
many roots will its corresponding quadratic equation have?
13-4 GRAPHIC SOLUTION OF A QUADRATIC-LINEAR SYSTEM
In Chapter 10 you learned how to solve a system of linear equations by graphing. For example, the graphic solution of the given system of linear equations is
shown below.
y 5 212x 1 4
y
Two points of
intersection
One point of
intersection
2x
–
1
y=
Since the point of intersection, (2, 3), is a solution of both equations, the common solution of
the system is x 2 and y 3.
A quadratic-linear system consists of a
quadratic equation and a linear equation. The
solution of a quadratic-linear system is the set
of ordered pairs of numbers that make both
equations true. As shown below, the line may
intersect the curve in two, one, or no points.
Thus the solution set may contain two ordered
pairs, one ordered pair, or no ordered pairs.
1
y 2x 1
O
1
(2, 3)
y=
– –1
2x+
4
x
No point of
intersection
526
Quadratic Relations and Functions
EXAMPLE 1
y x2 6x 6
Solve the quadratic-linear system graphically:
yx4
Solution (1) Draw the graph of y x2 6x 6.
The axis of symmetry of a parabola is x 2b
2a . Therefore, the axis of symme2(26)
try for the graph of y x2 6x 6 is x 2(1) or x 3. Make a table of
values using integral values of x less than 3 and greater than 3. Plot the
points associated with each pair (x, y) and join them with a smooth curve.
x2 6x 6
0
006
6
1
166
1
2
4 12 6
2
3
9 18 6
3
4
16 24 6
2
5
25 30 6
1
6
36 36 6
6
y
y
2
y=x
x
– 6x +
6
x
O
(2) On the same set of axes, draw the graph of y x 4. Make a table of values and plot the points.
2
4
44
0
6
(5, 1)
x
O
(2, –2)
4
24
–
2
x
4
=
04
y
0
y
– 6x +
y
2
x4
y=x
x
Graphic Solution of a Quadratic-Linear System
527
The line could also have been graphed by using the slope 5 1 or 11, and the
y-intercept, 4. Starting at the point (0, 4) move 1 unit up and 1 unit to
the right to locate a second point. Then again, move 1 unit up and 1 unit to
the right to locate a third point. Draw a line through these points.
(3) Find the coordinates of the points at which the graphs intersect. The
graphs intersect at (2, 2) and at (5, 1). Check each solution in each equation. Four checks are required in all. The checks are left for you.
Calculator (1) Enter the equations into the Y= menu.
Solution
6
x2
ENTER: Y X,T,,n
DISPLAY:
X,T,,n
6
X,T,,n
4
Plot1 Plot2 Plot3
\ Y 1 = X 2– 6 X + 6
\Y2= X–4
(2) Calculate the first intersection by choosing intersect from the CALC
menu. Accept Y1 X2 6X 6 as the first curve and Y2 X 4
as the second curve. Then press enter when the screen prompts “Guess?”.
ENTER:
2nd
CALC
5
ENTER
ENTER
ENTER
*
DISPLAY:
intersection
X= 2
Y = –2
The calculator displays the coordinates of the intersection point at the bottom of the screen.
Answer The solution set is {(2, 2), (5, 1)}.
*
(3) To calculate the second intersection point,
repeat the process from (2), but when the
screen prompts “Guess?” move the cursor
with the left and right arrow keys to the
approximate position of the second intersection point.
intersection
X= 5
Y= 1
528
Quadratic Relations and Functions
EXERCISES
Writing About Mathematics
1. What is the solution set of a system of equations when the graphs of the equations do not
intersect? Explain your answer.
2. Melody said that the equations y x2 and y 2 do not have a common solution even
before she drew their graphs. Explain how Melody was able to justify her conclusion.
Developing Skills
In 3–6, use the graph on the right to find
the common solution of the system.
y
y=x
2
3. y x2 2x 3
+ 2x –
y0
1
3
4. y x2 2x 3
O
y 3
–1
–1
5. y x2 2x 3
x
y 4
6. y x2 2x 3
y5
7. For what values of c do the equations y x2 2x 3 and y c have no points in common?
8. a. Draw the graph of y x2 4x 2, in the interval 1 x 5.
b. On the same set of axes, draw the graph of y x 2.
c. Write the coordinates of the points of intersection of the graphs made in parts a and b.
d. Check the common solutions found in part c in both equations.
In 9–16, find graphically and check the solution set of each system of equations.
9. y x2
yx2
11. y x2 2x 1
y 2x 5
13. y x2 8x 15
xy5
15. y x2 4x 1
y 2x 1
10. y x2 2x 4
yx
12. y 4x x2
yx4
14. y x2 6x 5
y3
16. y x2 x 4
2x y 2
529
Algebraic Solution of a Quadratic-Linear System
Applying Skills
17. When a stone is thrown upward from a height of 5 feet with an initial velocity of
48 feet per second, the height of the stone y after x seconds is given by the function
y 5 216x2 1 48x 1 5.
a. Draw a graph of the given function. Let each horizontal unit equal 21 second and each
vertical unit equal 5 feet. Plot points for 0, 12, 1, 32, 2, 52, and 3 seconds.
b. On the same set of axes, draw the graph of y 25.
c. From the graphs drawn in parts a and b, determine when the stone is at a height of
25 feet.
18. If you drop a baseball on Mars, the gravity accelerates the baseball at 12 feet per second
squared. Let’s suppose you drop a baseball from a height of 100 feet. A formula for the
height, y, of the baseball after x seconds is given by y 6x2 100.
a. On a calculator, graph the given function. Set Xmin to –10, Xmax to 10, Xscl to 1,
Ymin to 5, Ymax to 110, and Yscl to 10.
b. Graph the functions y 46 and y 0 as Y2 and Y3.
c. Using a calculator method, determine, to the nearest tenth of a second, when the baseball has a height of 46 feet and when the baseball hits the ground (reaches a height of
0 feet).
13-5 ALGEBRAIC SOLUTION OF A QUADRATIC-LINEAR SYSTEM
y
1
1
x2 –
O
x+
2
4x
+
2
1
1
3
y=
y=
In the last section, we learned to solve a
quadratic-linear system by finding the
points of intersection of the graphs. The
solutions of most of the systems that we
solved were integers that were easy to
read from the graphs. However, not all
solutions are integers. For example, the
graphs of y x2 4x 3 and y 5 12x 1 1
are shown at the right. They intersect in
two points. One of those points, (4, 3), has
integral coordinates and can be read easily
from the graph. However, the coordinates
of the other point are not integers, and we
are not able to identify the exact values of
x and y from the graph.
x
530
Quadratic Relations and Functions
In Chapter 10 we learned that a system of linear equations can be solved
by an algebraic method of substitution. This method can also be used for a
quadratic-linear system. The algebraic solution of the system graphed on the
previous page is shown in Example 1.
EXAMPLE 1
Solve algebraically and check:
y x2 4x 3
y 5 12x 1 1
y x2 4x 3
Solution (1) Since y is expressed in terms of x
in the linear equation, substitute
the expression 12x 1 1 for y in the
quadratic equation to form an
equation in one variable:
(2) To eliminate fractions as coefficients,
multiply both sides of the equation
by 2:
1
2x
1 1 x2 4x 3
2 A 12x 1 1 B 2(x2 4x 3)
x 2 2x2 8x 6
(3) Write the quadratic equation in
standard form:
0 2x2 9x 4
(4) Solve the quadratic equation by
factoring:
0 (2x 1)(x 4)
(5) Set each factor equal to 0 and solve
for x.
2x 1 0
2x 1
x 5 12
x40
x4
(6) Substitute each value of x in the
linear equation to find the
corresponding value of y:
y 12x 1 1
y 12x 1 1
y 12 A 12 B 1 1
y 12 (4) 1 1
y 14 1 1
y21
y 54
y3
(x, y) A 12, 54 B
(8) Check each ordered pair in each of the given equations:
(7) Write each solution as coordinates:
Check for x 21, y 54
y 12x 1 1
y x2 4x 3
5 ? 1 2
1
4 5 A2B 24A2B
5 ? 1
4542213
5
5
4 5 4✔
13
5 ? 1 1
4 5 2A2B 1
5 ? 1
45411
5
5
4 5 4✔
1
(x, y) (4, 3)
Algebraic Solution of a Quadratic-Linear System
531
Check for x 4, y 3
y x2 4x 3
y 12x 1 1
?
3 5 12 (4) 1 1
3 5 16 2 16 1 3
?
35211
3 53✔
353✔
3 5 (4) 2 2 4(4) 1 3
?
?
Answer U A 12, 54 B , (4, 3) V
EXAMPLE 2
The length of the longer leg of a right triangle is 2 units more than twice the
length of the shorter leg. The length of the hypotenuse is 13 units. Find the
lengths of the legs of the triangle.
Solution Let a the length of the longer leg and b the length of the shorter leg.
a2 b2 (13)2
a2 b2 169
a 2b 2
(1) Use the Pythagorean Theorem to
write an equation:
(2) Use the information in the first
sentence of the problem to write
another equation:
(3) Substitute the expression for a
from step 2 in the equation in
step 1:
(4) Square the binomial and write the
equation in standard form:
(5) Factor the left member of the
equation:
(6) Set each factor equal to 0 and solve
for b:
a2 b2 169
(2b 2)2 b2 169
4b2 8b 4 b2 169
5b2 8b 165 0
(5b 33)(b 5) 0
5b 33 0
5b 33
b
(7) For each value of b, find the value
of a. Use the linear equation in
step 2:
(8) Reject the negative values. Use the
pair of positive values to write the
answer.
233
5
b50
b5
b5
a 2b 2
a 2b 2
a 2 A 233
5 B 12
a 2(5) 2
256
a 5
Answer The lengths of the legs are 12 units and 5 units.
a 12
532
Quadratic Relations and Functions
EXERCISES
Writing About Mathematics
1. Explain why the equations y x2 and y 4 have no common solution in the set of real
numbers.
2. Explain why the equations x2 y2 49 and x 8 have no common solution in the set of
real numbers.
Developing Skills
In 3–20, solve each system of equations algebraically and check all solutions.
3. y x2 2x
yx
4. y x2 5
5. y x2 4x 3
yx5
6. y x 2x 1
2
yx3
9. x 2y 5
2
yx1
12. y 3x 8x 5
yx1
7. x y 9
8. x2 y 2
2
yx9
10. y 2x 6x 5
2
y x
11. y 2x2 2x 3
yx2
yx3
2
13. y x 3x 1
14. y x2 6x 8
xy3
y 5 13x 1 2
y 5 212x 1 2
15. x2 y2 25
16. x2 y2 100
2
x 2y 5
18. x y 40
2
2
y 2x 2
17. x2 y2 50
yx2
19. x y 20
2
2
xy
20. x2 y2 2
yx2
yx2
Applying Skills
21. A rectangular tile pattern consists of a large square, two
small squares, and a rectangle arranged as shown in the
diagram. The height of the small rectangle is 1 unit and
the area of the tile is 70 square units.
y
x
a. Write a linear equation that expresses the relationship
between the length of the sides of the quadrilaterals
that make up the tile.
b. Write a second-degree equation using the sum of the
areas of the quadrilaterals that make up the tile.
c. Solve algebraically the system of equations written in parts a and b.
d. What are the dimensions of each quadrilateral in the tile?
1
y
Chapter Summary
22. A doorway to a store is in the shape of an arch whose equation is
h 5 234x2 1 6x, where x represents the horizontal distance, in feet,
from the left end of the base of the doorway and h is the height, in
feet, of the doorway x feet from the left end of the base.
a. How wide is the doorway at its base?
b. What is the maximum height of the doorway?
533
h
(0, 0)
x
c. Can a box that is 6 feet wide, 6 feet long, and 5 feet high be moved through the doorway? Explain your answer.
23. The length of the diagonal of a rectangle is "85 meters. The length of the rectangle is 1
meter longer than the width. Find the dimensions of the rectangle.
24. The length of one leg of an isosceles triangle is "29 feet. The length of the altitude to the
base of the triangle is 1 foot more than the length of the base
a. Let a the length of the altitude to the base and b the distance from the vertex of a
base angle to the vertex of the right angle that the altitude makes with the base. Use
the Pythagorean Theorem to write an equation in terms of a and b.
b. Represent the length of the base in terms of b.
c. Represent the length of the altitude, a, in terms of b.
d. Solve the system of equations from parts a and c.
e. Find the length of the base and the length of the altitude to the base.
f. Find the perimeter of the triangle.
g. Find the area of the triangle.
CHAPTER SUMMARY
The equation y ax2 bx c, where a 0, is a quadratic function whose
domain is the set of real numbers and whose graph is a parabola. The axis of
symmetry of the parabola is the vertical line x 2b
2a . The vertex or turning point
of the parabola is on the axis of symmetry. If a 0, the parabola opens upward
and the y-value of the vertex is a minimum value for the range of the function.
If a 0, the parabola opens downward and the y-value of the vertex is a maximum value for the range of the function.
A quadratic-linear system consists of two equations one of which is an
equation of degree two and the other a linear equation (an equation of degree
one). The common solution of the system may be found by graphing the equations on the same set of axes or by using the algebraic method of substitution.
A quadratic-linear system of two equations may have two, one, or no common
solutions.
534
Quadratic Relations and Functions
The roots of the equation ax2 bx c 0 are the x-coordinates of
the points at which the function y ax2 bx c intersects the x-axis. The
real number k is a root of ax2 bx c 0 if and only if (x k) is a factor of
ax2 bx c.
When the graph of y x2 is translated by k units in the vertical direction,
the equation of the image is y x2 k. When the graph of y x2 is translated
k units in the horizontal direction, the equation of the image is y (x k)2.
When the graph of y x2 is reflected over the x-axis, the equation of the image
is y x2. The graph of y kx2 is the result of stretching the graph of y x2
in the vertical direction when k 1 or of compressing the graph of y x2 when
0 k 1.
VOCABULARY
13-1 Standard form • Polynomial equation of degree two • Quadratic
equation • Roots of an equation • Double root
13-2 Parabola • Second-degree polynomial function • Quadratic function
• Minimum • Turning point • Vertex • Axis of symmetry of a parabola
• Maximum
13-4 Quadratic-linear system
REVIEW EXERCISES
1. Explain why x y2 is not a function when the domain and range are the
set of real numbers.
2. Explain why x y2 is a function when the domain and range are the set of
positive real numbers.
In 3–6, the set of ordered pairs of a relation is given. For each relation, a. list the
elements of the domain, b. list the elements of the range, c. determine if the
relation is a function.
3. {(1, 3), (2, 2), (3, 1), (4, 0), (5, 1)}
4. {(1, 3), (1, 2), (1, 1), (1, 0), (1, 1)}
5. {(3, 9), (2, 4), (1, 1), (0, 0), (1, 1), (2, 4), (3, 9)}
6. {(0, 1), (1, 1), (2, 1), (3, 1), (4, 1)}
In 7–10, for each of the given functions:
a. Write the equation of the axis of symmetry.
b. Draw the graph.
c. Write the coordinates of the turning point.
Review Exercises
535
d. Does the function have a maximum or a minimum?
e. What is the range of the function?
7. y x2 6x 6
8. y x2 4x 1
9. f(x) x2 2x 6
10. f(x) x2 6x 1
In 11–16, solve each system of equations graphically and check the solution(s)
if they exist.
11. y x2 6
xy6
14. y x2 4x
xy4
12. y x2 2x 1
yx5
15. y 5 x2
y4
13. y x2 x 3
yx
16. y 2x x2
xy2
In 17–22, solve each system of equations algebraically and check the solutions.
17. x2 y 5
y 3x 1
20. y x2 6x 5
18. y x2 4x 9
y 1 2x
21. x2 y2 40
19. x2 2y 11
yx4
22. x2 y2 5
y 5 12x
23. Write an equation for the resulting function if the graph of y 5 x2 is
shifted 3 units up, 2.5 units to the left, and is reflected over the x-axis.
yx1
y 3x
24. The sum of the areas of two squares is 85. The length of a side of the
larger square minus the length of a side of the smaller square is 1. Find the
length of a side of each square.
25. Two square pieces are cut from a rectangular piece of carpet as shown in the diagram. The area of the original piece is 144
square feet, and the width of the small rectangle that is left is 2 feet. Find the dimensions of the original piece of carpet.
Exploration
Write the square of each integer from 2 to 20. Write the prime factorization of
each of these squares. What do you observe about the prime factorization of
each of these squares?
Let n be a positive integer and a, b, and c be prime numbers. If the prime
factorization of n 5 a3 3 b2 3 c4, is n a perfect square? Is "n rational or irrational? Express "n in terms of a, b, and c.
536
Quadratic Relations and Functions
CUMULATIVE REVIEW
CHAPTERS 1–13
Part I
Answer all questions in this part. Each correct answer will receive 2 credits. No
partial credit will be allowed.
1. When a 7, 5 2a2 equals
(1) 147
(2) 93
(4) 147
(3) 103
2. Which expression is rational?
(2) $13
(3) $28
(4) "0.4
3. When factored completely, 3x2 75 can be expressed as
(1) (3x 15)(x 5)
(3) 3x(x 5)(x 5)
(2) (x 5)(3x 15)
(4) 3(x 5)(x 5)
(1) p
4. When b2 4b is subtracted from 3b2 3b the difference is
(1) 3 7b
(2) 2b2 7b
(3) 2b2 7b
(4) 2b2 b
5. The solution set of the equation 0.5x 4 2x 0.5 is
(1) {3}
(2) {1.4}
(3) {30}
(4) {14}
6. Which of these represents the quadratic function y x 5 shifted 2 units
down and 4 units to the right?
(1) y (x 4)2 2
(3) y (x 4)2 7
2
(2) y (x 4) 3
(4) y (x 2)2 9
2
7. The slope of the line whose equation is x 2y 4 is
(1) 2
(3) 21
(2) 2
(4) 212
8. The solution set of the equation x2 7x 10 0 is
(1) {–2, 5}
(3) (x 2)(x 5)
(2) {2, 5}
(4) (x 2)(x 5)
9. Which of these shows the graph of a linear function intersecting the graph
of an absolute value function?
y
y
(1)
(2)
x
x
Cumulative Review
(3)
(4)
y
537
y
x
x
10. In ABC, mA 72 and mB 83. What is the measure of C?
(1) 155°
(2) 108°
(3) 97°
(4) 25°
Part II
Answer all questions in this part. Each correct answer will receive 2 credits.
Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit.
11. In a class of 300 students, 242 take math, 208 take science, and 183 take
both math and science. How many students take neither math nor science?
12. Each of the equal sides of an isosceles triangle is 3 centimeters longer than
the base. The perimeter of the triangle is 54 centimeters, what is the measure of each side of the triangle?
Part III
Answer all questions in this part. Each correct answer will receive 3 credits.
Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit.
13. The base of a right circular cylinder has a diameter of 5.00 inches. Sally
measured the circumference of the base of the cylinder and recorded it to
be 15.5 inches. What is the percent of error in her measurement? Express
your answer to the nearest tenth of a percent.
14. Solve the following system of equations and check your solutions.
y x2 3x 1
yx1
538
Quadratic Relations and Functions
Part IV
Answer all questions in this part. Each correct answer will receive 4 credits.
Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit.
15. Find to the nearest degree the measure of the acute angle that the graph
of y 2x 4 makes with the x-axis.
16. Jean Forester has a small business making pies and cakes. Today, she must
make at least 4 cakes to fill her orders and at least 3 pies. She has time to
make a total of no more than 10 pies and cakes.
a. Let x represent the number of cakes that Jean makes and y represent
the number of pies. Write three inequalities that can be used to represent the number of pies and cakes that she can make.
b. In the coordinate plane, graph the inequalities that you wrote and indicate the region that represents their common solution.
c. Write at least three ordered pairs that represent the number of pies and
cakes that Jean can make.

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project