1. No category

Name ———————————————————————
LESSON
10.1
Date ————————————
Study Guide
For use with pages 628 – 634
GOAL
Graph simple quadratic functions.
Vocabulary
A quadratic function is a nonlinear function that can be written in the
standard form y 5 ax 2 1 bx 1 c where a Þ 0.
Every quadratic function has a U-shaped graph called a parabola.
The most basic quadratic function in the family of quadratic functions,
called the parent quadratic function, is y 5 x 2.
LESSON 10.1
The lowest or highest point on a parabola is the vertex.
The line that passes through the vertex and divides the parabola into
two symmetric parts is called the axis of symmetry.
EXAMPLE 1
Graph y 5 ax 2 when ⏐a⏐ > 1
Graph y 5 26x 2. Compare the graph with the graph of y 5 x 2.
Solution
12
9
x
22
21
0
1
2
y
224
26
0
26
224
STEP 2
Plot the points from the table.
STEP 3
Draw a smooth curve through the points.
STEP 4
Compare the graphs of y 5 26x 2 and y 5 x 2.
Both graphs have the same vertex, (0, 0), and
the same axis of symmetry, x 5 0. However,
the graph of y 5 26x 2 is narrower than the
graph of y 5 x 2 and it opens down. This is
because the graph of y 5 26x 2 is a vertical
stretch (by a factor of 6) of the graph of y 5 x 2
and a reflection in the x-axis of the graph of y 5 x 2.
Algebra 1
Chapter 10 Resource Book
y
Make a table of values for y 5 26x 2.
y 5 x2
3
23
21
29
215
221
1
3
x
y 5 26x 2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
STEP 1
Name ———————————————————————
LESSON
10.1
Study Guide
Date ————————————
continued
For use with pages 628 – 634
EXAMPLE 2
Graph y 5 ax 2 when⏐a⏐< 1
2 2
Graph y 5 }
x . Compare the graph with the graph of y 5 x 2.
5
STEP 1
2
y
Make a table of values for y 5 }5 x 2.
35
x
210
25
0
5
10
y
40
10
0
10
40
25
2
y 5 5 x2
y 5 x2
Plot the points from the table.
STEP 3
Draw a smooth curve through the points.
STEP 4
Compare the graphs of y 5 }5 x 2 and y 5 x 2.
2
215
25
5
x
15
LESSON 10.1
STEP 2
Both graphs have the same vertex, (0, 0), and the same axis of symmetry,
2
x 5 0. However, the graph of y 5 }5 x 2 is wider than the graph of y 5 x 2.
This is because the graph of y 5 }5 x 2 is a vertical shrink 1 by a factor of }5 2
2
2
of the graph of y 5 x 2.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
EXAMPLE 3
Graph y 5 ax 2 1 c
Graph y 5 3x 2 2 1. Compare the graph with the graph of y 5 x 2.
STEP 1
Make a table of values for y 5 3x 2 2 1.
y
x
22
21
0
1
2
10
y
11
2
21
2
11
6
y 5 x2
2
STEP 2
Plot the points from the table.
STEP 3
Draw a smooth curve through the points.
STEP 4
Compare the graphs of y 5 3x 2 2 1 and
y 5 x 2. Both graphs open up and have the same axis of symmetry, x 5 0.
However, the graph of y 5 3x 2 2 1 is narrower and has a lower vertex than
the graph of y 5 x 2. This is because the graph of y 5 3x 2 2 1 is a vertical
stretch and a vertical translation of the graph of y 5 x 2.
23
21
22
1
3
y5
3x 2 2
x
1
Exercises for Examples 1, 2, and 3
Graph the function. Compare the graph with the graph of y 5 x 2.
1
1
3. y 5 2} x 2
3
1
1
6. y 5 2} x 2 2 1
2
1. y 5 28x 2
2.
y 5 }7 x 2
4. y 5 x 2 2 3
5.
y 5 }4 x 2 1 2
Algebra 1
Chapter 10 Resource Book
13
Answer Key
Lesson 10.1
Study Guide
1.
Both graphs have the same vertex, (0, 0), and the same axis of symmetry, x 5 0. However, the graph of y
5 28x 2 is narrower than the graph of y 5 x 2 and it opens down. This is because the graph of y 5 28x 2 is a
vertical stretch (by a factor of 8) of the graph of y 5 x 2 and a reflection in the x-axis of the graph of y 5 x 2.
2.
3.
oth graphs have the same vertex, (0, 0),
B
and the same axis of
symmetry, x 5 0.
However, the graph of
1
y 5 ​ }7 ​ x 2 is wider than the graph of y 5 x 2. This is because the graph of
1
1
y5}
​ 7 ​ x 2 is a vertical shrink 1​ by a factor of }
​ 7 ​ 2​of the graph of y 5 x 2.
Both graphs have the same vertex, (0, 0), and the same axis of symmetry, x 5 0. However, the graph
1
of y 5 2​ }3 ​ x 2 is narrower than the graph of y 5x 2
1
1
and it opens down. This is because the graph of y 5 2​ }3 ​ x 2 is a vertical shrink 1​ by a factor of }
​ 3 ​ 2​of
the graph of y 5 x 2 and a reflection in the x-axis of the graph of y 5 x 2.
4.
Both graphs have the same axis of symmetry,
x 5 0, and both open up. However, the graph of
y 5 x 2 2 3 has a lower vertex than the graph of
y 5 x 2. This is because the graph of y 5 x 2 2 3 is a vertical translation of the graph of y 5 x 2.
5.
Both graphs open up, and have the same axis of symmetry, x 5 0. However, the graph of
1
y5}
​ 4 ​ x 2 1 2 is wider than the graph of y 5 x 2, and has a higher vertex. This is because the
1
1
graph of y 5 }
​ 4 ​ x 2 1 2 is a vertical shrink 1​ by a factor of }
​ 4 ​ 2​and a vertical translation of the
graph of y 5 x 2.
6.
1
Both graphs have the same axis of symmetry, x 5 0. However, the graph of y 5 2​ }2 ​ x 2 2 1 is
wider than the graph of y 5 x 2, opens down and has a lower vertex. This is because the graph of
y 5 2​ }2 ​ x 2 2 1 is a vertical shrink 1​ by a factor of }
​ 2 ​ 2​and a vertical translation of the graph of y 5 x 2.
1
1
Name ———————————————————————
LESSON
10.2
Date ————————————
Study Guide
For use with pages 635 – 640
GOAL
Graph general quadratic functions.
Vocabulary
For y 5 ax 2 1 bx 1 c, the y-coordinate of the vertex is the minimum
value of the function if a > 0 and the maximum value of the function
if a < 0.
EXAMPLE 1
Find the axis of symmetry and the vertex
Consider the function y 5 3x2 2 18x 1 11.
a. Find the axis of symmetry of the graph of the function.
b. Find the vertex of the graph of the function.
Solution
a. For the function y 5 3x 2 2 18x 1 11, a 5 3 and b 5 218.
(218)
b
x 5 2}
5 2}
53
2a
2(3)
Substitute 3 for a and 218 for b.
Then simplify.
b
b. The x-coordinate of the vertex is 2}, or 3. To find the y-coordinate,
2a
substitute 3 for x in the function and find y.
y 5 3(3)2 218(3) 1 11 5 216
Substitute 3 for x. Then simplify.
The vertex is (3, 216).
EXAMPLE 2
Find the minimum or maximum value
LESSON 10.2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The axis of symmetry is x 5 3.
Tell whether the function f (x) 5 x 2 1 14x 2 3 has a minimum value or
a maximum value. Then find the minimum or maximum value.
Solution
Because a 5 1 and 1 > 0, the parabola opens up and the function has a minimum
value. To find the minimum value, find the vertex.
b
14
b
x 5 2}
5 2}
5 27
2a
2(1)
The x-coordinate is 2}
.
2a
f (27) 5 (27)2 1 14(27) 2 3 5 252
Substitute 27 for x. Then simplify.
The minimum value of the function is f (x) 5 252.
Algebra 1
Chapter 10 Resource Book
27
Name ———————————————————————
LESSON
10.2
Study Guide
Date ————————————
continued
For use with pages 635 – 640
Exercises for Examples 1 and 2
Find the axis of symmetry and the vertex of the graph of the function.
1. y 5 5x 2 1 20x 1 9
1
2. y 5 } x 2 2 4x 2 19
3
1
3. Tell whether the function f(x) 5 } x 2 2 8x 1 13 has a minimum value or a
2
maximum value. Then find the minimum value or maximum value.
EXAMPLE 3
Graph y 5 ax 2 1 bx 1 c
1 2
Graph y 5 }
x 2 2x 1 3.
5
Solution
STEP 1
Determine whether the parabola opens up or down.
Because a > 0, the parabola opens up.
STEP 2
Find and draw the axis of symmetry:
(22)
y
21 }5 2
STEP 3
14
x55
10
Find and plot the vertex.
b
LESSON 10.2
The x-coordinate of the vertex is 2}
,
2a
or 5. To find the y-coordinate, substitute
5 for x in the function and simplify.
(0, 3)
(10, 3)
(1, 1.2)
22
1
y 5 }5(5)2 2 2(5) 1 3 5 22
(9, 1.2)
6
10
14 x
(5, 22)
So, the vertex is (5, 22).
STEP 4
Plot two points. Choose two x-values less than the x-coordinate of the vertex.
Then find the corresponding y-values.
x
0
1
y
3
1.2
STEP 5
Reflect the points plotted in Step 4 in the axis of symmetry.
STEP 6
Draw a parabola through the plotted points.
Exercise for Example 3
4. Graph the function f (x) 5 x 2 2 4x 1 7. Label the vertex and axis of symmetry.
28
Algebra 1
Chapter 10 Resource Book
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
b
x 5 2}
5 2}
5 5.
2a
1
Answer Key
Lesson 10.2
Study Guide
1. x 5 22: (22, 211) 2. x 5 6: (6, 231)
3. minimum value; 219
4.
Name ———————————————————————
LESSON
10.3
Date ————————————
Study Guide
For use with pages 643–651
GOAL
Solve quadratic equations by graphing.
Vocabulary
A quadratic equation is an equation that can be written in the
standard form ax 2 1 bx 1 c 5 0 where a Þ 0 and a is called
the leading coefficient.
EXAMPLE 1
Solve a quadratic equation having two solutions
Solve x 2 1 5x 5 14 by graphing.
x 5 27 y
Solution
STEP 1
Write the equation in standard form.
x 2 1 5x 5 14
x 2 1 5x 2 14 5 0
STEP 2
22
22
x52
x
26
Write original equation.
Subtract 14 from
each side.
210
214
Graph the function y 5 x 2 1 5x 2 14.
The solutions of the equation x 2 1 5x 5 14
are 27 and 2.
You can check 27 and 2 in the original equation.
CHECK
x 2 1 5x 5 14
x 2 1 5x 5 14
(27) 1 5(27) 0 14
(2) 1 5(2) 0 14
2
14 5 14 ✓
EXAMPLE 2
2
14 5 14 ✓
Write original equation.
Substitute for x.
Simplify. Each solution checks.
Solve a quadratic equation having one solution
Solve x 2 1 25 5 10x by graphing.
Solution
y
STEP 1
Write the equation in standard form.
x 2 1 25 5 10x
x 2 10x 1 25 5 0
2
STEP 2
Write original equation.
Subtract 10x from
each side.
Graph the function y 5 x 2 2 10x 1 25.
The x-intercept is 5.
LESSON 10.3
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The x-intercepts are 27 and 2.
10
6
x55
2
2
6
10
x
The solution of the equation x 2 1 25 5 10x is 5.
Algebra 1
Chapter 10 Resource Book
41
Name ———————————————————————
LESSON
10.3
Study Guide
Date ————————————
continued
For use with pages 643–651
EXAMPLE 3
Solve a quadratic equation having no solution
Solve x2 1 11 5 5x by graphing.
Solution
STEP 1
x 2 1 11 5 5x
x2 2 5x 1 11 5 0
STEP 2
y
Write the equation in standard form.
Write original equation.
10
Subtract 5x from each side.
6
Graph the function y 5 x 2 2 5x 1 11.
2
The graph has no x-intercepts.
22
The equation x2 1 11 5 5x has no solution.
2
6
x
Exercises for Examples 1, 2, and 3
Solve the equation by graphing.
1. x 2 5 2x 1 15
2. x 2 1 4 5 24x
EXAMPLE 4
Find the zeros of a quadratic function
Find the zeros of f(x) 5 x 2 2 10x 1 24.
Solution
Graph the function f (x) 5 x 2 2 10x 1 24.
The x-intercepts are 4 and 6.
y
10
The zeros of the function are 4 and 6.
6
x56
2
LESSON 10.3
x54
Exercises for Example 4
Find the zeros of the function.
4. f (x) 5 x 2 2 4
5. f (x) 5 x 2 1 5x 2 14
42
Algebra 1
Chapter 10 Resource Book
10
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3. x 2 1 6x 5 24
Answer Key
Lesson 10.3
Study Guide
}
}
1. 23, 5 2. 22 3. 23 1 ​Ï 5 ​, 23 2 ​Ï 5 ​ 4. 22, 2 5. 27, 2
Name ———————————————————————
LESSON
LESSON 10.4
10.4
Date ————————————
Study Guide
For use with pages 652– 658
GOAL
EXAMPLE 1
Solve a quadratic equation by finding square roots.
Solve quadratic equations
Solve the equation.
a. x 2 2 7 5 9
b.
11y 2 5 11
c. z 2 1 13 5 5
Solution
a. x 2 2 7 5 9
Write original equation.
x 2 5 16
Add 7 to each side.
}
x 5 6Ï16
Take square roots of each side.
5 64
Simplify.
The solutions are 24 and 4.
b. 11y 2 5 11
Write original equation.
y2 5 1
Divide each side by 11.
}
y 5 6Ï 1
Take square roots of each side.
5 61
Simplify.
c. z 2 1 13 5 5
Write original equation.
z 5 28
Subtract 13 from each side.
Negative real numbers do not have real square roots. So, there is
no solution.
2
EXAMPLE 2
Take square roots of a fraction
Solve 9m 2 5 169.
Solution
9m2 5 169
Write original equation.
169
m2 5 }
9
Divide each side by 9.
}
Ï
169
m56 }
9
Take square roots of each side.
13
m 5 6}
3
Simplify.
13
13
and }
.
The solutions are 2}
3
3
52
Algebra 1
Chapter 10 Resource Book
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The solutions are 21 and 1.
Name ———————————————————————
LESSON
10.4
Study Guide
Date ————————————
continued
For use with pages 652– 658
Approximate solutions of a quadratic equation
Solve 2x 2 1 5 5 15. Round the solutions to the nearest hundredth.
Solution
2x 2 1 5 5 15
Write original equation.
2x 2 5 10
LESSON 10.4
EXAMPLE 3
Subtract 5 from each side.
x2 5 5
Divide each side by 2.
}
x 5 ± Ï5
Take square roots of each side.
x ≈ ±2.24
Use a calculator. Round to the nearest hundredth.
The solutions are about 22.24 and about 2.24.
Exercises for Examples 1, 2, and 3
Solve the equation.
1. w 2 2 9 5 0
2.
4r 2 2 7 5 9
3. 5s 2 1 13 5 9
4. 36x 2 5 121
5.
16m2 1 81 5 81
6. 4q2 2 225 5 0
Solve the equation. Round the solutions to the nearest hundredth.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
7. 7x 2 2 8 5 13
EXAMPLE 4
8.
26y 2 1 15 5 215
9. 4z 2 1 7 5 12
Solve a quadratic equation
Solve 3(x 1 3)2 5 39. Round the solutions to the nearest hundredth.
Solution
3(x 1 3)2 5 39
Write original equation.
(x 1 3)2 5 13
Divide each side by 3.
}
x 1 3 5 ± Ï13
Take square roots of each side.
}
x 5 23 ± Ï13
Subtract 3 from each side.
}
}
The solutions are 23 1 Ï13 ø 0.61 and 23 2 Ï13 ø 26.61.
Exercises for Example 4
Solve the equation.
10. 5(x 2 1)2 5 40
11.
2( y 1 4)2 5 18
12. 4(z 2 5)2 5 32
Algebra 1
Chapter 10 Resource Book
53
Answer Key
Lesson 10.4
Study Guide
15 15
11 11
1. 23, 3 2. 22, 2 3. no solution 4. 2​ } ​ , ​ } ​ 5. 0 6. 2​ } ​ , }
​ ​ 7. 21.73, 1.73
16 16
2 2
8. 22.24, 2.24 9. 21.12, 1.12 10. 21.83, 3.83 11. 21, 27 12. 2.17, 7.83
Name ———————————————————————
LESSON
10.5
Date ————————————
Study Guide
For use with pages 662–668
GOAL
Solve quadratic equations by completing the square.
Vocabulary
For an expression of the form x 2 1 bx, you can add a constant c to
the expression so that the expression x 2 1 bx 1 c is a perfect square
trinomial. This process is called completing the square.
EXAMPLE 1
Complete the square
LESSON 10.5
Find the value of c that makes the expression x 2 1 7x 1 c a perfect
square trinomial. Then write the expression as the square of a binomial.
Solution
Find the value of c. For the expression to be a perfect square trinomial,
c needs to be the square of half the coefficient of x.
c 5 1 }2 2 5 }
4
7 2
STEP 2
49
Find the square of half the coefficient of x.
Write the expression as a perfect square trinomial. Then write the expression
as the square of a binomial.
49
x 2 1 7x 1 c 5 x 2 1 7x 1 }
4
5 1 x 1 }2 2
7 2
EXAMPLE 2
49
Substitute }
for c.
4
Square of a binomial.
Solve a quadratic equation
Solve x 2 1 14x 5 213 by completing the square.
Solution
x 2 1 14x 5 213
x 2 1 14x 1 (7)2 5 213 1 72
Write original equation.
Add 1 }
, or 7 2, to each side.
22
14 2
(x 1 7)2 5 213 1 72
Write left side as the square of a binomial.
(x 1 7)2 5 36
Simplify the right side.
x17566
x 5 27 6 6
Take square roots of each side.
Subtract 7 from each side.
The solutions of the equation are 27 1 6 5 21 and 27 2 6 5 213.
64
Algebra 1
Chapter 10 Resource Book
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
STEP 1
Name ———————————————————————
LESSON
10.5
Study Guide
Date ————————————
continued
For use with pages 662–668
EXAMPLE 3
Solve a quadratic equation in standard form
Solve 3x2 1 18x 2 9 5 0 by completing the square. Round your
solutions to the nearest hundredth.
Solution
3x2 1 18x 2 9 5 0
Write original equation.
3x 1 18x 5 9
2
Add 9 to each side.
x2 1 6x 5 3
Divide each side by 3.
Add 1 }2 2 , or 32, to each side.
6 2
x2 1 6x 1 32 5 3 1 32
Write left side as the square of a binomial.
}
x 1 3 5 Ï12
Take square roots of each side.
}
x 5 23 6 Ï12
}
Subtract 3 from each side.
}
The solutions are 23 1 Ï12 ø 0.46 and 23 2 Ï12 ø 26.46.
LESSON 10.5
(x 1 3)2 5 12
Exercises for Examples 1, 2, and 3
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Find the value of c that makes the expression a perfect square
trinomial. Then write the expression as the square of a binomial.
1. x 2 2 9x 1 c
2. x 2 1 11x 1 c
3. x 2 2 16x 1 c
Solve the equation by completing the square. Round your solutions to
the nearest hundredth if necessary.
4. q2 2 8q 5 7
5. r 2 1 12r 5 23
6. 2s 2 2 28s 1 8 5 0
Algebra 1
Chapter 10 Resource Book
65
Answer Key
Lesson 10.5
Study Guide
81
9 2
121
11 2
1. ​ } ​ ; ​ x 2 }
​ 2 ​ ​ 2. }
​ 4 ​ ; ​ x 1 }
​ 2 ​ ​ 3. 64; (x 2 8)2 4. 20.8, 8.8
4
1
2
1
2
5. 20.26, 211.74 6. 13.71, 0.29
Name ———————————————————————
LESSON
10.6
Date ————————————
Study Guide
For use with pages 671– 676
GOAL
Solve quadratic equations using the quadratic formula.
Vocabulary
By completing the square for the quadratic equation ax2 1 bx 1 c 5 0,
}
2b 6 Ïb2 2 4ac
you can develop a formula, x 5 }}
, that gives the
2a
solutions of any quadratic equation in standard form. This formula
is called the quadratic formula.
EXAMPLE 1
Solve a quadratic equation
Solve 5x 2 2 3 5 4x.
Solution
5x 2 2 3 5 4x
Write original equation.
5x 2 2 4x 2 3 5 0
Write in standard form.
}
2b 6 Ï b2 2 4ac
x 5 }}
2a
Quadratic formula
LESSON 10.6
Substitute values in the quadratic formula:
a 5 5, b 5 24, and c 5 23.
}
4 6 Ï 76
10
5}
Simplify.
}
4 1 Ï 76
}
4 2 Ï 76
The solutions are }
ø 1.27 and }
ø 20.47.
10
10
Exercises for Example 1
Use the quadratic formula to solve the equation. Round your solutions
to the nearest hundredth, if necessary.
1. x 2 2 12x 2 14 5 0
2. 5y 2 2 7 5 11y
3. 9z 2 1 3z 5 5
74
Algebra 1
Chapter 10 Resource Book
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}}
2(24) 6 Ï (24)2 2 4(5)(23)
5 }}}
2(5)
Name ———————————————————————
LESSON
10.6
Study Guide
Date ————————————
continued
For use with pages 671– 676
EXAMPLE 2
Use the quadratic formula
Retirement Savings For the period 1995–2005, the amount of dollars
invested in an individual’s retirement account can be modeled by the
function y 5 30x2 2 24x 1 15,500 where x is the number of years since
1995. In what year was $17,000 invested?
Solution
y 5 30x 2 2 24x 1 15,500
Write function.
17,000 5 30x 2 2 24x 1 15,500
Substitute 17,000 for y.
0 5 30x 2 2 24x 2 1500
Write in standard form.
}}
2(224) 6 Ï (224)2 2 4(30)(21500)
2(30)
x 5 }}}
Substitute values in the quadratic formula:
a 5 30, b 5 224, and c 5 21500.
}
24 6 Ï 180,576
5 }}
60
Simplify.
}
}
24 1 Ï 180,576
24 2 Ï 180,576
The solutions are }}
ø 7 and }}
ø 27.
60
60
The year when $17,000 is invested is about 7 years after 1995, or 2002.
Choose a solution method
Tell what method you would use to solve the quadratic equation.
Explain your choice(s).
a. 3x 2 1 13x 5 11
b.
x 2 1 8x 5 7
c. 4x 2 2 25 5 0
Solution
LESSON 10.6
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
EXAMPLE 3
a. The quadratic equation cannot be factored easily, and completing
the square will result in many fractions. So, the equation can be
solved using the quadratic formula.
b. The quadratic equation can be solved by completing the square
because the equation can be rewritten in the form ax 2 1 bx 1 c 5 0
where a 5 1 and b is an even number.
c. The quadratic equation can be solved using square roots because
the equation can be written in the form x 2 5 d.
Exercises for Examples 2 and 3
4. In Example 2, find the year when $18,000 was invested.
Tell what method you would use to solve the quadratic equation.
Explain your choice(s).
5. x 2 1 11x 5 0
6.
23x 2 1 19x 5 27
7. 4x 2 1 16x 5 12
Algebra 1
Chapter 10 Resource Book
75
Answer Key
Lesson 10.6
Study Guide
1. 21.07, 13.07 2. 22.72, 0.52 3. 20.9, 0.6 4. 2005 5. factor or complete the square
6. quadratic formula 7. complete the square
Name ———————————————————————
LESSON
10.7
Date ————————————
Study Guide
For use with pages 677–683
GOAL
Use the value of the discriminant.
Vocabulary
In the quadratic formula, the expression b 2 2 4ac is called the
discriminant of the associated equation ax 2 1 bx 1 c 5 0.
EXAMPLE 1
EXAMPLE 2
Use the discriminant
Equation
ax 2 1 bx 1 c
Discriminant
b 2 2 4ac
Number of
solutions
a.
9x 2 1 30x 1 25 5 0
30 2 2 4(9)(25) 5 0
One solution
b.
7x 2 2 4x 1 6 5 0
(24)2 2 4(7)(6) 5 2152
No solution
c.
4x 2 2 8x 1 3 5 0
(28)2 2 4(4)(3) 5 16
Two solutions
Find the number of solutions
Solution
STEP 1
Write the equation in standard form.
16x 2 1 49 5 56x
16x 2 2 56x 1 49 5 0
STEP 2
Write equation.
Subtract 56x from each side.
Find the value of the discriminant.
b2 2 4ac 5 (256)2 2 4(16)(49)
50
Substitute 16 for a, 256 for b,
and 49 for c.
Simplify.
LESSON 10.7
The discriminant is zero, so the equation has one solution.
Exercises for Examples 1 and 2
Tell whether the equation has two solutions, one solution, or
no solution.
1. 2x 2 1 x 5 21
2. 4x 2 1 5x 1 2 5 0
3. 25x 2 1 4 5 20x
84
Algebra 1
Chapter 10 Resource Book
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Tell whether the equation 16x2 1 49 5 56x has two solutions, one
solution, or no solution.
Name ———————————————————————
LESSON
10.7
Study Guide
Date ————————————
continued
For use with pages 677–683
EXAMPLE 3
Find the number of x-intercepts
Find the number of x-intercepts of the graph of y 5 x2 2 12x 1 36.
Solution
Find the number of solutions of the equation 0 5 x 2 2 12x 1 36.
b2 2 4ac 5 (212)2 2 4(1)(36)
50
Substitute 1 for a, 212 for b, and 36 for c.
Simplify.
The discriminant is zero, so the equation has one solution. This means that the graph
of y 5 x 2 2 12x 1 36 has one x-intercept.
CHECK
You can use a graphing calculator to check your answer. Notice that the
graph of y 5 x 2 2 12x 1 36 intercepts the x-axis once.
Find the number of x-intercepts of the graph.
4. y 5 7x2 2 14x
5. y 5 x2 1 7x 1 13
6. y 5 4x2 2 12x 1 9
LESSON 10.7
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Exercises for Example 3
Algebra 1
Chapter 10 Resource Book
85
Answer Key
Lesson 10.7
Study Guide
1. two solutions 2. no solution 3. one solution 4. 2 5. 0 6. 1
Name ———————————————————————
LESSON
LESSON 10.8
10.8
Date ————————————
Study Guide
For use with pages 684–693
GOAL
EXAMPLE 1
Compare linear, exponential, and quadratic models.
Choose functions using sets of ordered pairs
Use a graph to tell whether the ordered pairs represent a linear
function, an exponential function, or a quadratic function.
a. (22, 216), ( 21, 215), (0, 212), (1, 27), (2, 0)
b. (22, 1), ( 21, 3), (0, 5), (1, 7), (2, 9)
1
1
, 21, }5 2, (0, 1), (1, 5), (2, 25)
1 22, }
25 2 1
c.
Solution
b.
y
y
27
7
21
26
5
15
210
3
9
1
3
22
22
x
21
Quadratic function
EXAMPLE 2
c.
y
9
2
Linear function
1
x
23
21
1
x
Exponential function
Identify functions using differences or ratios
Use differences or ratios to tell whether the table of values represents
a linear function, an exponential function, or a quadratic function.
Solution
x
21
0
1
2
y
1
3
9
27
3
Ratios: }1 5 3
3
3
The table represents an exponential function.
98
Algebra 1
Chapter 10 Resource Book
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
a.
Name ———————————————————————
LESSON
10.8
Study Guide
Date ————————————
continued
For use with pages 684–693
1. Tell whether the ordered pairs represent a linear function, a quadratic function,
or an exponential function: (21, 26), (0, 24), (1, 0), (2, 6).
2. Tell whether the table represents a linear function, a quadratic function, or an
exponential function.
EXAMPLE 3
x
0
1
2
3
y
26
3
12
21
LESSON 10.8
Exercises for Examples 1 and 2
Write an equation for a function
Tell whether the table of values represents a linear function, an
exponential function, or a quadratic function. Then write an equation
for the function.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
STEP 1
Determine which type of function the values in the table represent.
x
21
0
1
2
3
y
7
5
3
1
21
First differences: 2 2 2 2 2 2 2 2
The table of values represents a linear function because the first differences
are equal.
STEP 2
Write an equation for the linear function. The equation has the form
y 5 mx 1 b. When x 5 0, y 5 5, so b 5 5. Find m by substituting any
two points into the slope formula.
y
527
0 2 (21)
7
22
m5}5}
5 22
1
5
The equation is y 5 22x 1 5.
3
CHECK
Plot the ordered pairs from the table.
Then graph y 5 22x 1 5 to see that
the graph passes through the plotted
points.
1
21
21
1
x
Exercises for Example 3
Tell whether the table of values represents a linear function, an
exponential function, or a quadratic function. Then write an equation
for the function.
3.
x
21
0
1
2
y
12
6
2
0
4.
x
22
21
y
0.0625 0.125 0.25 0.5 1
0
1
2
Algebra 1
Chapter 10 Resource Book
99
Answer Key
Lesson 10.8
Study Guide
1. quadratic function 2. linear function 3. quadratic function: y 5 x 2 2 5x 1 6
4. exponential function: y 5 (0.25)(2) x