Quadratic Functions and Transformations€¦ · 4-1 Review Parent Quadratic Function The parent...

Name Class Date

4-1 Review

Parent Quadratic Function The parent quadratic function is y = x2.

Substitute 0 for x in the function to get y = 0. The vertex of the parent quadratic function is (0, 0).

A few points near the vertex are:

The graph is symmetrical about the line x = 0. This line is the axis of symmetry.

Vertex Form of a Quadratic Function The vertex form of a quadratic function is y = a(x − h)2 + k. The graph of this function is a transformation of the graph of the parent quadratic function y = x2. The vertex of the graph is (h, k). If a = 1, you can graph the function by sliding the graph of the parent function h units along the x-axis and k units along the y-axis.

What is the graph of y = (x + 3)2 + 2? What are the vertex and axis of symmetry of the function?

Step 1 Write the function in vertex form: y = 1[x − (−3)]2 + 2

Step 2 Find the vertex: h = −3, k = 2. The vertex is (−3, 2).

Step 3 Find the axis of symmetry. Since the vertex is (−3, 2), the graph is symmetrical about the line x = −3. The axis of symmetry is x = −3.

Step 4 Because a = 1, you can graph this function by sliding the graph of the parent function −3 units along the x-axis and 2 units along the y-axis. Plot a few points near the vertex to help you sketch the graph.

Prentice Hall Algebra 2 • Teaching Resources Copyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

9

Quadratic Functions and Transformations

x –3 –2 –1 1 2 3

y 9 4 1 1 4 9

x –5 –4 –3 –2 –1

y 6 3 2 3 6

Name Class Date

4-1 Review (continued)

If a ≠ 1, the graph is a stretch or compression of the parent function by a factor of | a |.

0 < | a | < 1 | a | > 1 The graph is a vertical compression The graph is a vertical stretch

of the parent function. of the parent function

What is the graph of y = 2(x + 3)2 + 2?

Step 1 Write the function in vertex form: y = 2[x − (−3)]2 + 2

Step 2 The vertex is (−3, 2). Step 3 The axis of symmetry is x = −3. Step 4 Because a = 2, the graph of this function is a vertical

stretch by 2 of the parent function. In addition to sliding the graph of the parent function 3 units left and 2 units up, you must change the shape of the graph. Plot a few points near the vertex to help you sketch the graph.

Exercises Graph each function. Identify the vertex and axis of symmetry.

1. y = (x − 1)2 + 3 2. y = (x + 4)2 − 2 3. y = (x + 2)2 + 1

4. y = 2(x − 1) 2 + 3 5. y = 12

(x + 4) 2 − 2 6. y = 0.9(x + 2)2 + 1


10


x –5 –4 –3 –2 –1

y 10 4 2 4 10

Name Class Date

4-2 Review

• The graph of a quadratic function, y = ax2 + bx + c, where a ≠ 0, is a parabola.

• The axis of symmetry is the line .

• The x-coordinate of the vertex is . The y-coordinate of the vertex is

, or the y-value when .

• The y-intercept is (0, c).

What is the graph of y = 2x2 − 8x + 5?

Find the equation of the axis of symmetry.

x-coordinate of vertex: 2

Find the y-value when x = 2.

= 8 – 16 + 5 = –3

y-coordinate of vertex: −3 The vertex is (2, −3). y-intercept: (0, 5) The y-intercept is at (0, c) = (0, 5).

Because a is positive, the graph opens upward, and the vertex is at the bottom of the graph. Plot the vertex and draw the axis of symmetry. Plot (0, 5) and its corresponding point on the other side of the axis of symmetry.

Exercises Graph each parabola. Label the vertex and the axis of symmetry.

1. y = −3x2 + 6x − 9 2. y = −x2 − 8x − 15

3. y = 2x2 − 8x + 1 4. y = −2x2 − 12x − 7


19

Standard Form of a Quadratic Function

Name Class Date

4-2 Review (continued)

What is the vertex form of y = 3x2 − 24x + 50?

y = ax2 + bx + c y = 3x2 − 24x + 50 Verify that the equation is in standard form. b = −24, a = 3 Find b and a.

x -coordinate = For an equation in standard form, the x-coordinate of the vertex

can be found by using .

Substitute.

= 4 Simplify.

y-coordinate = 3(4)2 − 24(4) + 50 Substitute 4 into the standard form to find the y-coordinate.

= 2 Simplify.

y = 3(x − 4)2 + 2 Substitute 4 for h and 2 for k into the vertex form.

Once the conversion to vertex form is complete, check by multiplying.

y = 3(x2 − 8x + 16) + 2 y = 3x2 − 24x + 50 The result is the standard form of the equation.

Exercises Write each function in vertex form. Check your answers.

5. y = x2 − 2x − 3 6. y = −x2 + 4x + 6 7. y = x2 + 3x − 10

8. y = x2 − 9x 9. y = x2 + x 10. y = x2 + 5x + 4

11. y = 4x2 + 8x − 3 12. y = x2 + 9x 13. y = −2x2 + 2x + 1

Write each function in standard form.

14. y = (x − 3)2 + 1 15. y = 2(x − 1)2 − 3 16. y = −3(x + 4)2 + 1


20


• Standard form of a quadratic function is y = ax2 + bx + c. Vertex form of a quadratic function is y = a(x − h)2 + k.

• For a parabola in vertex form, the coordinates of the vertex are (h, k).

Name Class Date

4-3 Review

A parabola contains the points (0, −2), (−1, 5), and (2, 2). What is the equation of this parabola in standard form?

If the parabola y = ax2 + bx + c passes through the point (x, y), the coordinates of the point must satisfy the equation of the parabola. Substitute the (x, y) values into y = ax2 + bx + c to write a system of equations.

First, use the point (0, −2). y = ax2 + bx + c Write the standard form.

−2 = a(0)2 + b(0) + c Substitute.

−2 = c Simplify.

Use the point (−1, 5) next. 5 = a(−1)2 + b(−1) + c Substitute.

5 = a − b + c Simplify.

Finally, use the point (2, 2). 2 = a(2)2 + b(2) + c Substitute.

2 = 4a + 2b + c Simplify. Because c = −2, the resulting system has two variables. Simplify the equations above.

a − b = 7 4a + 2b = 4

Use elimination to solve the system and obtain a = 3, b = –4, and c = −2. Substitute these values into the standard form y = ax2 + bx + c.

The equation of the parabola that contains the given points is y = 3x2 − 4x − 2.

Exercises Find an equation in standard form of the parabola passing through the given points.

1. (0, −1), (1, 5), (−1, −5) 2. (0, 4), (−1, 9), (2, 0)

3. (0, 1), (1, 4), (3, 22) 4. (1, −1), (−2, 20), (2, 0)

5. (−1, −5), (0, −1), (2, 1) 6. (1, 3), (−2, −3), (−1, 3)


29

Modeling With Quadratic Functions

Three non-collinear points, no two of which are in line vertically, are on the graph of exactly one quadratic function.

Name Class Date

4-3

Review (continued)

A soccer player kicks a ball of the top of a building. His friend records the height of the ball at each second. Some of her data appears in the table.

a. What is a quadratic model for these data? b. Use the model to complete the table.

Use the points (0, 112), (1, 192), and (5, 192) to find the quadratic model. Substitute the (t, h) values into h = at2 + bt + c to write a system of equations.

(0, 112): 112 = a(0)2 + b(0) + c c = 112

(1, 192): 192 = a(1)2 + b(1) + c a + b + c = 192

(5, 192): 192 = a(5)2 + b(5) + c 25a + 5b + c = 192

Use c = 112 and simplify the equations to obtain a system with just two variables.

a + b = 80

25a + 5b = 80

Use elimination to solve the system. The quadratic model for the data is

h = −16t2 + 96t + 112

Now use this equation to complete the table for the t-values 2, 3, 4, 6, and 7.

t = 2: h = −16(2)2 + 96(2) + 112 = −64 + 192 + 112 = 240

t = 3: h = −16(3)2 + 96(3) + 112 = −144 + 288 + 112 = 256

t = 4: h = −16(4)2 + 96(4) + 112 = −256 + 384 + 112 = 240

t = 6: h = −16(6)2 + 96(6) + 112 = −576 + 576 + 112 = 112

t = 7: h = −16(7)2 + 96(7) + 112 = −784 + 672 + 112 = 0

Exercise 7. The number n of Brand X shoes in stock at the beginning of month t in a store

follows a quadratic model. In January (t = 1), there are 36 pairs of shoes; in March (t = 3), there are 52 pairs; and in September, there are also 52 pairs. a. What is the quadratic model for the number n of pairs of shoes at the

beginning of month t? b. How many pairs are in stock in June?


30

Modeling With Quadratic Functions

Time (s) Height (ft)

0 112

1 192

2

3

4

5 192 6 7


0 112

1 192

2 240 3 256

4 240 5 192

6 112

7 0

Name Class Date

4-4 What is 6x2 − 5x − 4 in factored form?

a = 6, b = –5, and c = –4 Find a, b, and c; they are the coefficients of each term.

ac = −24 and b = –5 We are looking for factors with product ac and sum b.

The factors 3 and −8 are the combination whose sum is –5.

Rewrite the middle term using the factors you found. Find common factors by grouping the terms in pairs. Rewrite using the Distributive Property. Check (3x − 4)(2x + 1) You can check your answer by multiplying the factors together.

6x2 + 3x − 8x − 4 6x2 − 5x − 4

Remember that not all quadratic expressions are factorable.

Factoring Quadratic Expressions

Reteaching

Factors of –24 1, –24 –1,–24 2,–12 –2,12 3, –8 –3,8 4, –6 –4, 6 Sum of factors –23 23 –10 10 –5 5 –2 2

Exercises

Factor each expression.

1. x2 + 6x + 8 2. x2 − 4x + 3

3. 2x2 − 6x + 4 4. 2x2 − 11x + 5

5. 2x2 − 7x − 4 6. 4x2 + 16x + 15

7. x2 − 5x − 14 8. 7x2 − 19x − 6

9. x2 − x − 72 10. 2x2 + 9x + 7

11. x2 + 12x + 32 12. 4x2 − 28x + 49

13. x2 − 3x − 10 14. 2x2 + 9x + 4

15. 9x2 − 6x + 1 16. x2 − 10x + 9

17. x2 + 4x − 12 18. x2 + 7x + 10

19. x2 − 8x + 12 20. 2x2 − 5x − 3

21. x2 − 6x + 5 22. 3x2 + 2x − 8


39

Name Class Date

4-4 Reteaching (continued)

What is 25x2 − 20x + 4 in factored form?

There are three terms. Therefore, the expression may be a perfect square trinomial. a2 = 25x2 and b2 = 4 Find a2 and b2.

a = 5x and b = 2 Take square roots to find a and b.

Check that the choice of a and b gives the correct middle term.

2ab = 2 • 5x • 2 = 20x Write the factored form. a2 − 2ab + b2 = (a − b)2

25x2 − 20x + 4 = (5x − 2)2

Check (5x − 2)2 You can check your answer by multiplying the factors together. (5x − 2)(5x − 2) Rewrite the square in expanded form.

25x2 − 10x − 10x + 4 Distribute. 25x2 − 20x + 4 Simplify.

Exercises Factor each expression.

23. x2 − 12x + 36 24. x2 + 30x + 225 25. 9x2 − 12x + 4

26. x2 − 64 27. 9x2 − 42x + 49 28. 25x2 − 1

29. 27x2 − 12 30. 49x2 + 42x + 9 31. 16x2 − 32x + 16

32. 9x2 − 16 33. 8x2 − 18 34. 81x2 + 126x + 49

35. 125x2 − 100x + 20 36. −x2 + 196 37. −16x2 − 24x − 9


40

Factoring Quadratic Expressions • a2 + 2ab + b2 = (a + b)2 Factoring perfect square trinomials a2 − 2ab + b2 = (a − b)2 • a2 − b2 = (a + b)(a − b) Factoring a difference of two squares

Name Class Date

4-5 Review

There are several ways to solve quadratic equations. If you can factor the quadratic expression in a quadratic equation written in standard form, you can use the Zero-Product Property.

If ab = 0 then a = 0 or b = 0.

What are the solutions of the quadratic equation 2x2 + x = 15?

2x2 + x = 15 Write the equation.

2x2 + x − 15 = 0 Rewrite in standard form, ax2 + bx + c = 0. (2x − 5)(x + 3) = 0 Factor the quadratic expression (the nonzero side).

2x − 5 = 0 or x + 3 = 0 Use the Zero-Product Property. 2x = 5 or x = −3 Solve for x.

or x = −3

Check the solutions:

15 x = −3: 2(−3)2 + (−3) 15

15 18 − 3 15

15 = 15 15 = 15

Both solutions check. T e solutions are x = and x = −3.

Exercises Solve each equation by factoring. Check your answers.

1. x2 − 10x + 16 = 0 2. x2 + 2x = 63 3. x2 + 9x = 22

4. x2 − 24x + 144 = 0 5. 2x2 = 7x + 4 6. 2x2 = −5x + 12

7. x2 − 7x = −12 8. 2x2 + 10x = 0 9. x2 + x = 2

10. 3x2 − 5x + 2 = 0 11. x2 = −5x − 6 12. x2 + x = 20


49

Quadratic Equations

Name Class Date

4-5

Review (continued)

Some quadratic equations are dif cult or impossible to solve by factoring. You can use a graphing calculator to find the points where the graph of a function intersects the x-axis. At these points f(x) = 0, so x is a zero of the function.

The values r1 and r2 are the zeros of the function y = (x − r1)(x − r2). The graph of the function intersects the x-axis at x = r1, or (r1, 0), and x = r2, or (r2, 0).

What are the solutions of the quadratic equation 3x2 = 2x + 7?

Step 1 Rewrite the equation in standard form, ax2 + bx + c = 0. 3x2 − 2x − 7 = 0

Step 2 Enter the equation as Y1 in your calculator.

Step 3 Graph Y1. Choose the standard window and see if the zeros of the function Y1 are visible on the screen. If they are not visible, zoom out and determine a better viewing window. In this case, the zeros are visible in the standard window.

Step 4 Use the ZERO option in the CALC feature. For the first zero, choose bounds of −2 and −1 and a guess of −1.5. The screen display gives the first zero as x = −1.230139.

Similarly, the screen display gives the second zero as x = 1.8968053.

The solutions to two decimal places are x = −1.23 and x = 1.90.

Exercises Solve the equation by graphing. Give each answer to at most two decimal places.

13. x2 = 5 14. x2 = 5x + 1

15. x2 + 7x = 3 16. x2 + x = 5

17. x2 + 3x + 1 = 0 18. x2 = 2x + 4

19. 3x2 − 5x + 9 = 8 20. 4 = 2x2 + 3x

21. x2 − 6x = –7 22. −x2 = 8x + 8


50

Quadratic Equations

Name Class Date

4-6 Review

Completing a perfect square trinomial allows you to factor the completed trinomial as the square of a binomial.

Start with the expression x2 + bx. Add 2

2b⎛ ⎞⎜ ⎟⎝ ⎠

. Now the expression is 2

2

2bx bx ⎛ ⎞+ + ⎜ ⎟⎝ ⎠

,

which can be factored into the square of a binomial: 2 2

2

2 2b bx bx x⎛ ⎞ ⎛ ⎞+ + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

.

To complete the square for an expression ax2 + abx, first factor out a. Then find the value that completes the square for the factored expression.

What value completes the square for −2x2 + 10x?

Think Write

−2x2 + 10x = −2(x2 − 5x)

5 52 2 2b −= = −

2

2 25 252 5 2 52 4

x x x x⎡ ⎤⎛ ⎞ ⎛ ⎞− − + − = − − +⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

252

2x⎛ ⎞− −⎜ ⎟⎝ ⎠

25 2524 2

⎛ ⎞− = −⎜ ⎟⎝ ⎠

Exercises What value completes the square for each expression?

1. x2 + 2x 2. x2 − 24x 3. x2 + 12x

4. x2 − 20x 5. x2 + 5x 6. x2 − 9x

7. 2x2 − 24x 8. 3x2 + 12x 9. −x2 + 6x

10. 5x2 + 80x 11. −7x2 + 14x 12. −3x2 − 15x


59

Completing the Square

Write the expression in the form a(x2

+ bx).

Find .2b

Add 2

2b⎛ ⎞⎜ ⎟⎝ ⎠

to the inner expression to

complete the square.

Factor the perfect square trinomial.

Find the value that completes the square.

Name Class Date

4-6

Review (continued)

You can easily graph a quadratic function if you first write it in vertex form. Complete the square to change a function in standard form into a function in vertex form.

What is y = x2 − 6x + 14 in vertex form?

Think Write

x2 − 6x

6 32 2b −= = −

x2 − 6x + (−3)2 = x2 − 6x + 9

y = x2 − 6x + 9 + 14 − 9

y = (x − 3)2 + 14 − 9

y = (x − 3)2 + 5

Exercises Rewrite each equation in vertex form.

13. y = x2 + 4x + 3 14. y = x2 − 6x + 13

15. y = 2x2 + 4x − 10 16. y = x2 − 2x − 3

17. y = x2 + 8x + 13 18. y = –x2 − 6x − 4

19. y = −x2 + 10x − 18 20. y = x2 + 2x − 8

21. y = 2x2 + 4x − 3 22. y = 3x2 − 12x + 8


60

Completing the Square

Write an expression using the terms that contain x.

Find .2b

Add 2

2b⎛ ⎞⎜ ⎟⎝ ⎠

to the expression to

complete the square.

Subtract 9 from the expression so that the equation is unchanged.


Add the remaining constant terms.

Name Class Date

4-7 Review

You can solve some quadratic equations by factoring or completing the square. You can solve any quadratic equation ax2 + bx + c = 0 by using the Quadratic Formula:

Notice the ± symbol in the formula. Whenever b2 − 4ac is not zero, the Quadratic Formula will result in two solutions.

What are the solutions for 2x2 + 3x = 4? Use the Quadratic Formula.

2x2 + 3x − 4 = 0 Write the equation in standard form: ax2 + bx + c = 0 a = 2; b = 3; c = −4 a is the coefficient of x2, b is the coefficient of x, c is

the constant term.

Write the Quadratic Formula.

Substitute 2 for a, 3 for b, and −4 for c.

Simplify.

Write the solutions separately.

Check your results on your calculator. Replace x in the original equation with

and . Both values

for x give a result of 4. The solutions check.

What are the solutions for each equation? Use the Quadratic Formula.

1. −x2 + 7x − 3 = 0 2. x2 + 6x = 10

3. 2x2 = 4x + 3 4. 4x2 + 81 = 36x

5. 2x2 + 1 = 5 − 7x 6. 6x2 − 10x + 3 = 0


69

The Quadratic Formula

Exercises

4-7

Quadratic Formula Review (continued)

There are three possible outcomes when you take the square root of a real number n:

two real values (one positive and one negative)

one real value (0)

no real values

Now consider the quadratic formula: . The value under

the radical symbol determines the number of real solutions that exist for the equation ax2 + bx + c = 0:

two real solutions

one real solution

no real solutions

What is the number of real solutions of −3x2 + 7x = 2?

−3x2 + 7x = 2

−3x2 + 7x − 2 = 0 Write in standard form.

a = −3, b = 7, c = −2 Find the values of a, b, and c.

b2 − 4ac Write the discriminant.

(7)2 − 4(−3)(−2) Substitute for a, b, and c.

49 − 24 Simplify.

25

The discriminant, 25, is positive. The equation has two real roots.

Exercises What is the value of the discriminant and what is the number of real

solutions for each equation?

The value under the radical, b2 – 4ac, is called the discriminant.

7. x2 + x − 42 = 0 8. −x2 + 13x − 40 = 0 9. x2 + 2x + 5 = 0

10. x2 = 18x − 81 11. −x2 + 7x + 44 = 0 12.

13. 2x2 + 7 = 5x 14. 4x2 + 25x = 21 15. x2 + 5 = 3x

16. 17. 18.


70

Name Class Date

4-8 Review

• A complex number consists of a real part and an imaginary part. It is written in the form a + bi, where a and b are real numbers.

• and

• When adding or subtracting complex numbers, combine the real parts and then combine the imaginary parts.

• When multiplying complex numbers, use the Distributive Property or FOIL.

What is (3 − i) + (2 + 3i)?

(3 − i) + (2 + 3i)

= Circle real parts. Put a square around imaginary parts.

= (3 + 2) + (−1 + 3)i Combine.

= 5 + 2i Simplify.

What is the product (7 − 3i)(−4 + 9i)?

Use FOIL to multiply:

(7 − 3i)(−4 + 9i) = 7(−4) + 7(9i) + (−3i)(−4) + (−3i)(9i)

(7 − 3i)(−4 + 9i) = −28 + 63i + 12i − 27i2

First = 7(−4) = −28 + 75i − 27i2

Outer = 7(9i) You can simplify the expression by substituting −1 for i2.

Inner = (−3i)(−4) (7 − 3i)(−4 + 9i) = −28 + 75i − 27(−1)

Last = (−3i)(9i) = −1 + 75i

Exercises Simplify each expression.

1. 2i + (−4 − 2i) 2. (3 + i)(2 + i) 3. (4 + 3i)(1 + 2i)

4. 3i(1 − 2i) 5. 3i(4 − i) 6. 3 − (−2 + 3i) + (−5 + i)

7. 4i(6 − 2i) 8. (5 + 6i) + (−2 + 4i) 9. 9(11 + 5i)


79

Complex Numbers

Name Class Date

4-8

Review (continued)

• The complex conjugate of a complex number a + bi is the complex number a − bi. • (a + bi)(a − bi) = a2 + b2

• To divide complex numbers, use complex conjugates to simplify the denominator.

What is the quotient

The complex

conjugate of 2 − i is 2 + i.

Multiply both numerator and denominator 2 + i.

Use FOIL to multiply the numerators.

Simplify the denominator. (a + bi)(a − bi) = a2 + b2

Substitute −1 for i2.

Simplify.

Write as a complex number a + bi.

Exercises Find the complex conjugate of each complex number.

10. 1 − 2i 11. 3 + 5i 12. i

13. 3 − i 14. 2 + 3i 15. −5 − 2i

Write each quotient as a complex number.

16. 17. 18.

19. 20. 21.

Prentice Hall Algebra 2 • Teaching Resources

Complex Numbers

Name Class Date

4-9 Review

You used graphing and substitution to solve systems of linear equations. You can use these same methods to solve systems involving quadratic equations.

What is the solution of the system of equations?

y = 2x − 3 Write one equation.

x2 − 2x− 8 = 2x − 3 Substitute x2 − 2x − 8 for y in the linear equation.

x2 − 4x − 5 = 0 Write in standard form.

(x + 1)(x − 5) = 0 Factor the quadratic expression.

x + 1 =0 or x − 5 = 0 Use the Zero-Product Property.

x = −1 or x = 5 Solve for x.

Because the solutions to the system of equations are ordered pairs of the form (x, y), solve for y by substituting each value of x into the linear equation. You can use either equation, but the linear equation is easier.

x = −1: y = 2x − 3 = 2(−1) − 3 = −5 → (−1, −5)

x = 5: y = 2x − 3 = 2(5) − 3 = 7 → (5, 7)

The solutions are (−1, −5) and (5, 7). Check these by graphing the system and identifying the points of intersection.

Exercises Solve each system.

1. 2.

3. 4.

5. 6.


89

Quadratic Systems

Name Class Date

4-9

Review (continued)

To solve a system of linear inequalities, you graph each inequality and find the region where the graphs overlap. You can also use this technique to solve a system of quadratic inequalities.

What is the solution of this system of inequalities?

Step 1 Graph the equation y = −x2 + 4x. Use a dashed boundary line because the points on the curve are not part of the solution. Choose a point on one side of the curve and check if it satisfies the inequality.

y < −x2 + 4x 0 −(2)2 + 4(2) Check the point (2, 0). 0 < 4 The inequality is true.

Points below the curve satisfy the inequality, so shade that region.

Step 2 Graph the equation y = x2 − 2x − 8. Use a dashed boundary line because the points on the curve are not part of the solution. Choose a point on one side of the curve and check if it satisfies the inequality.

y > x2 − 2x − 8 0 (2)2 − 2(2) − 8 Check the point (2, 0). 0 > −8 The inequality is true.

Points above the curve satisfy the inequality, so shade that region.

Step 3 The solution to the system of both inequalities is the set of points satisfying both inequalities. In other words, the solution is the region where the graphs overlap. The region contains no boundary points.

Exercises Solve each system by graphing.

7. 8.

9. 10.


90

Quadratic Systems

Prentice Hall Algebra 2 • Teaching ResourcesCopyright © by Pearson Education, Inc., or its affiliates. All Rights Reserved.

9

Name Class Date

Parent Quadratic Function

Th e parent quadratic function is y 5 x2.

Substitute 0 for x in the function to get y 5 0. Th e vertex of the parent quadratic function is (0, 0).

A few points near the vertex are:

Th e graph is symmetrical about the line x 5 0. Th is line is the axis of symmetry.

Vertex Form of a Quadratic Function

Th e vertex form of a quadratic function is y 5 a(x 2 h)2 1 k.Th e graph of this function is a transformation of the graph of the parent quadratic function y 5 x2. Th e vertex of the graph is (h, k). If a 5 1, you can graph the function by sliding the graph of the parent function h units along the x-axis and k units along the y-axis.

Problem

What is the graph of y 5 (x 1 3)2 1 2? What are the vertex and axis of symmetry of the function?

Step 1 Write the function in vertex form: y 5 1fx 2 (23)g2 1 2

Step 2 Find the vertex: h 5 23, k 5 2. Th e vertex is (23, 2).

Step 3 Find the axis of symmetry. Since the vertex is (23, 2), the graph is symmetrical about the line x 5 23. Th e axis of symmetry is x 5 23.

Step 4 Because a 5 1, you can graph this function by sliding the graph of the parent function 23 units along the x-axis and 2 units along the y-axis. Plot a few points near the vertex to help you sketch the graph.

4-1 ReteachingQuadratic Functions and Transformations

x

y

y � x2

x � 0

Vertex (0, 0)

Axis of symmetry

Slide h units

Slidek units

y � x2

y � a(x�h)2�k

x

y

Slide �3 units

Slide2 units

y � x2

y � (x�3)2 � 2

x

y

x

y

22

3

21

6

23

2

24

3

25

6

x

y

2

4

3

9

1

1

21

1

22

4

23

9


10

Name Class Date

If a 2 1, the graph is a stretch or compression of the parent function by a factor of u a u .

0 , u a u , 1 u a u . 1 Th e graph is a vertical Th e graph is a vertical compression of the parent function. stretch of the parent function

Problem

What is the graph of y 5 2(x 1 3)2 1 2?

Step 1 Write the function in vertex form: y 5 2fx 2 (23)g2 1 2

Step 2 Th e vertex is (23, 2).

Step 3 Th e axis of symmetry is x 5 23.

Step 4 Because a 5 2, the graph of this function is a vertical stretch by 2 of the parent function. In addition to sliding the graph of the parent function 3 units left and 2 units up, you must change the shape of the graph. Plot a few points near the vertex to help you sketch the graph.

Exercises

Graph each function. Identify the vertex and axis of symmetry.

1. y 5 (x 2 1)2 1 3 2. y 5 (x 1 4)2 2 2 3. y 5 (x 1 2)2 1 1

4. y 5 2(x 2 1)2 1 3 5. y 512(x 1 4)2 2 2 6. y 5 0.9(x 1 2)2 1 1



y � x2

y � ax2 x

y

Verticalcompression y � ax2 y � x2

x

y

Verticalstretch

Slide �3 units

Slide2 units

y � x2

x

yy � 2(x�3)2 � 2

Stretchvertically

x

y

22

4

21

10

23

2

24

4

25

10

(1, 3); x 5 1

(1, 3); x 5 1

(24, 22); x 5 24

(24, 22); x 5 24

(22, 1); x 5 22

(22, 1); x 5 22

xO

y

�2 2

2468

4 6

xO

y

�2�4�6�8�2

246

xO

y

�2 2�4�6

468

xO

y

2�2 4 6

2468

xO

y

�2�4�6�8�2

246

xO

y

�2 2�4�6

2468


19

Name Class Date

• Th e graph of a quadratic function, y 5 ax2 1 bx 1 c, where a 2 0, is a parabola.

• Th e axis of symmetry is the line x 5 2 b2a.

• Th e x-coordinate of the vertex is 2 b2a. Th e y-coordinate of the vertex is

y 5 f Q2 b2aR , or the y-value when x 5 2 b

2a.

• Th e y-intercept is (0, c).

Problem

What is the graph of y 5 2x2 2 8x 1 5?

x 5 2

b2a

52(28)

2(2)5

8

45 2 Find the equation of the axis of symmetry.

x-coordinate of vertex: 2 2

b2a

f Q2 b

2aR 5 f (2) 5 2(2)2 2 8(2) 1 5 Find the y-value when x 5 2.

5 8 2 16 1 5

5 23

y-coordinate of vertex: 23 The vertex is (2, 23).

y-intercept: (0, 5) The y-intercept is at (0, c) 5 (0, 5).

Because a is positive, the graph opens upward, and the vertex is at the bottom of the graph. Plot the vertex and draw the axis of symmetry. Plot (0, 5) and its corresponding point on the other side of the axis of symmetry.

Exercises

Graph each parabola. Label the vertex and the axis of symmetry.

1. y 5 23x2 1 6x 2 9 2. y 5 2x2 2 8x 2 15

3. y 5 2x2 2 8x 1 1 4. y 5 22x2 2 12x 2 7

4-2 ReteachingStandard Form of a Quadratic Function

y

4 5

�3�2

Ox

1

32

4

65

(2, �3)

x � 2

(4 , 5)(0, 5)

(1, �6)

(2, �9)(0, �9)

x � 14�4

yO x

�4

�8

3

y

�12

�18

3 x

�6

(�8, �15)

(�4, 1)

(0, �15)

x � �4�9

O

4

�4 4 8�4

�8

xy

x = 2(2, �7)

(0, 1) (4, 1)

O6

�4 4�8�6

x

y12

(0, �7)(�6, �7)

(�3, 11)x = �3


20

Name Class Date

• Standard form of a quadratic function is y 5 ax2 1 bx 1 c. Vertex form of a

quadratic function is y 5 a(x 2 h)2 1 k.

• For a parabola in vertex form, the coordinates of the vertex are (h, k).

Problem

What is the vertex form of y 5 3x2 2 24x 1 50?

y 5 ax2 1 bx 1 c

y 5 3x2 2 24x 1 50 Verify that the equation is in standard form.

b 5 224, a 5 3 Find b and a.

x-coordinate 5 2 b

2a For an equation in standard form, the x-coordinate of the vertex

can be found by using x 5 2 b2a.

5 2 2242(3)

Substitute.

5 4 Simplify.

y-coordinate 5 3(4)2 2 24(4) 1 50 Substitute 4 into the standard form to fi nd the y-coordinate.

5 2 Simplify.

y 5 3(x 2 4)2 1 2 Substitute 4 for h and 2 for k into the vertex form.

Once the conversion to vertex form is complete, check by multiplying.

y 5 3(x2 2 8x 1 16) 1 2

y 5 3x2 2 24x 1 50

Th e result is the standard form of the equation.

Exercises

Write each function in vertex form. Check your answers.

5. y 5 x2 2 2x 2 3 6. y 5 2x2 1 4x 1 6 7. y 5 x2 1 3x 2 10

8. y 5 x2 2 9x 9. y 5 x2 1 x 10. y 5 x2 1 5x 1 4

11. y 5 4x2 1 8x 2 3 12. y 534x2 1 9x 13. y 5 22x2 1 2x 1 1

Write each function in standard form.

14. y 5 (x 2 3)2 1 1 15. y 5 2(x 2 1)2 2 3 16. y 5 23(x 1 4)2 1 1



y 5 (x 2 1)2 2 4 y 5 2(x 2 2)2 1 10 y 5 Qx 1 32R2 2 49

4

y 5 Qx 2 92R2 2 81

4 y 5 Qx 1 12R2 2 1

4 y 5 Qx 1 52R2 2 9

4

y 5 4(x 1 1)2 2 7 y 5 34 (x 1 6)2 2 27 y 5 22Qx 2 1

2R2 1 32

y 5 x2 2 6x 1 10 y 5 2x2 2 4x 2 1 y 5 23x2 2 24x 2 47


29

Name Class Date

4-3 Reteaching Modeling With Quadratic Functions

Th ree non-collinear points, no two of which are in line vertically, are on the graph of exactly one quadratic function.

Problem

A parabola contains the points (0, 22), (21, 5), and (2, 2). What is the equation of this parabola in standard form?

If the parabola y 5 ax2 1 bx 1 c passes through the point (x, y), the coordinates of the point must satisfy the equation of the parabola. Substitute the (x, y) values into y 5 ax2 1 bx 1 c to write a system of equations.

First, use the point (0, 22). y 5 ax2 1 bx 1 c Write the standard form.

22 5 a(0)2 1 b(0) 1 c Substitute.

22 5 c Simplify.

Use the point (21, 5) next. 5 5 a(21)2 1 b(21) 1 c Substitute.

5 5 a 2 b 1 c Simplify.

Finally, use the point (2, 2). 2 5 a(2)2 1 b(2) 1 c Substitute.

2 5 4a 1 2b 1 c Simplify.

Because c 5 22, the resulting system has two variables. Simplify the equations above.

a 2 b 5 7

4a 1 2b 5 4

Use elimination to solve the system and obtain a 5 3, b 5 24, and c 5 22. Substitute these values into the standard form y 5 ax2 1 bx 1 c.

Th e equation of the parabola that contains the given points is y 5 3x2 2 4x 2 2.

Exercises

Find an equation in standard form of the parabola passing through the given points.

1. (0, 21), (1, 5), (21, 25) 2. (0, 4), (21, 9), (2, 0)

3. (0, 1), (1, 4), (3, 22) 4. (1, 21), (22, 20), (2, 0)

5. (21, 25), (0, 21), (2, 1) 6. (1, 3), (22, 23), (21, 3)

y 5 x2 1 5x 2 1 y 5 x2 2 4x 1 4

y 5 2x2 1 x 1 1 y 5 2x2 2 5x 1 2

y 5 2x2 1 3x 2 1 y 5 22x2 1 5


30

Name Class Date

4-3 Reteaching (continued) Modeling With Quadratic Functions

Problem

A soccer player kicks a ball off the top of a building. His friend records the height of the ball at each second. Some of her data appears in the table. a. What is a quadratic model for these data? b. Use the model to complete the table.

Use the points (0, 112), (1, 192), and (5, 192) to fi nd the quadratic model. Substitute the (t, h) values into h 5 at2 1 bt 1 c to write a system of equations.

(0, 112): 112 5 a(0)2 1 b(0) 1 c c 5 112

(1, 192): 192 5 a(1)2 1 b(1) 1 c a 1 b 1 c 5 192

(5, 192): 192 5 a(5)2 1 b(5) 1 c 25a 1 5b 1 c 5 192

Use c 5 112 and simplify the equations to obtain a system with just two variables.

a 1 b 5 80

25a 1 5b 5 80

Use elimination to solve the system. Th e quadratic model for the data is

h 5 216t2 1 96t 1 112

Now use this equation to complete the table for the t-values 2, 3, 4, 6, and 7.

t 5 2: h 5 216(2)2 1 96(2) 1 112 5 264 1 192 1 112 5 240

t 5 3: h 5 216(3)2 1 96(3) 1 112 5 2144 1 288 1 112 5 256

t 5 4: h 5 216(4)2 1 96(4) 1 112 5 2256 1 384 1 112 5 240

t 5 6: h 5 216(6)2 1 96(6) 1 112 5 2576 1 576 1 112 5 112

t 5 7: h 5 216(7)2 1 96(7) 1 112 5 2784 1 672 1 112 5 0

Exercise

7. Th e number n of Brand X shoes in stock at the beginning of month t in a store follows a quadratic model. In January (t 5 1), there are 36 pairs of shoes; in March (t 5 3), there are 52 pairs; and in September, there are also 52 pairs.

a. What is the quadratic model for the number n of pairs of shoes at the beginning of month t?

b. How many pairs are in stock in June?


0

1

2

3

4

5

6

7

112

192

192


0

1

2

3

4

5

6

7

112

192

240

256

240

192

112

0

n 5 2t2 1 12t 1 25

61


39

Name Class Date

Problem

What is 6x2 2 5x 2 4 in factored form?

a 5 6, b 5 25, and c 5 24 Find a, b, and c; they are the coeffi cients of each term.

ac 5 224 and b 5 25 We are looking for factors with product ac and sum b.

Th e factors 3 and 28 are the combination whose sum is 25.

6x2 1 3x 2 8x 2 4 Rewrite the middle term using the factors you found.

3x(2x 1 1) 2 4(2x 1 1) Find common factors by grouping the terms in pairs.

(3x 2 4)(2x 1 1) Rewrite using the Distributive Property.

Check (3x 2 4)(2x 1 1) You can check your answer by multiplying the factors together.

6x2 1 3x 2 8x 2 4

6x2 2 5x 2 4

Remember that not all quadratic expressions are factorable.

Exercises


1. x2 1 6x 1 8 2. x2 2 4x 1 3

3. 2x2 2 6x 1 4 4. 2x2 2 11x 1 5

5. 2x2 2 7x 2 4 6. 4x2 1 16x 1 15

7. x2 2 5x 2 14 8. 7x2 2 19x 2 6

9. x2 2 x 2 72 10. 2x2 1 9x 1 7

11. x2 1 12x 1 32 12. 4x2 2 28x 1 49

13. x2 2 3x 2 10 14. 2x2 1 9x 1 4

15. 9x2 2 6x 1 1 16. x2 2 10x 1 9

17. x2 1 4x 2 12 18. x2 1 7x 1 10

19. x2 2 8x 1 12 20. 2x2 2 5x 2 3

21. x2 2 6x 1 5 22. 3x2 1 2x 2 8

4-4 ReteachingFactoring Quadratic Expressions

3 3

Factors of 224

Sum of factors

1, 224

223

21, 24

23

2, 212

210

22, 12

10

3, 28

25

23, 8

5

4, 26

22

24, 6

2

(x 1 4)(x 1 2)

2(x 2 2)(x 2 1)

(2x 1 1)(x 2 4)

(x 1 2)(x 2 7)

(x 2 9)(x 1 8)

(x 1 4)(x 1 8)

(x 2 5)(x 1 2)

(3x 2 1)(3x 2 1)

(x 2 3)(x 2 1)

(2x 2 1)(x 2 5)

(2x 1 5)(2x 1 3)

(7x 1 2)(x 2 3)

(2x 1 7)(x 1 1)

(2x 2 7)(2x 2 7)

(2x 1 1)(x 1 4)

(x 2 1)(x 2 9)

(x 1 6)(x 2 2)

(x 2 6)(x 2 2)

(x 1 5)(x 1 2)

(2x 1 1)(x 2 3)

(x 2 1)(x 2 5) (3x 2 4)(x 1 2)


40

Name Class Date

• a2 1 2ab 1 b2 5 (a 1 b)2 Factoring perfect square trinomials

a2 2 2ab 1 b2 5 (a 2 b)2

• a2 2 b2 5 (a 1 b)(a 2 b) Factoring a difference of two squares

Problem

What is 25x2 2 20x 1 4 in factored form?

Th ere are three terms. Th erefore, the expression may be a perfect square trinomial.

a2 5 25x2 and b2 5 4 Find a2 and b2.

a 5 5x and b 5 2 Take square roots to fi nd a and b.

Check that the choice of a and b gives the correct middle term.

2ab 5 2 ? 5x ? 2 5 20x

Write the factored form.

a2 2 2ab 1 b2 5 (a 2 b)2

25x2 2 20x 1 4 5 (5x 2 2)2

Check (5x 2 2)2 You can check your answer by multiplying the factors together.

(5x 2 2)(5x 2 2) Rewrite the square in expanded form.

25x2 2 10x 2 10x 1 4 Distribute.

25x2 2 20x 1 4 Simplify.

Exercises


23. x2 2 12x 1 36 24. x2 1 30x 1 225 25. 9x2 2 12x 1 4

26. x2 2 64 27. 9x2 2 42x 1 49 28. 25x2 2 1

29. 27x2 2 12 30. 49x2 1 42x 1 9 31. 16x2 2 32x 1 16

32. 9x2 2 16 33. 8x2 2 18 34. 81x2 1 126x 1 49

35. 125x2 2 100x 1 20 36. 2x2 1 196 37. 216x2 2 24x 2 9


Factoring Quadratic Expressions

(x 2 6)2

(x 1 8)(x 2 8)

3(3x 1 2)(3x 2 2)

(3x 1 4)(3x 2 4)

5(5x 2 2)2

(x 1 15)2

(3x 2 7)2

(7x 1 3)2

2(2x 1 3)(2x 2 3)

2(x 1 14)(x 2 14)

(3x 2 2)2

(5x 1 1)(5x 2 1)

16(x 2 1)2

(9x 1 7)2

2(4x 1 3)2


49

Name Class Date

Th ere are several ways to solve quadratic equations. If you can factor the quadratic expression in a quadratic equation written in standard form, you can use the Zero-Product Property.

If ab 5 0 then a 5 0 or b 5 0.

Problem

What are the solutions of the quadratic equation 2x2 1 x 5 15?

2x2 1 x 5 15 Write the equation.

2x2 1 x 2 15 5 0 Rewrite in standard form, ax2 1 bx 1 c 5 0.

(2x 2 5)(x 1 3) 5 0 Factor the quadratic expression (the nonzero side).

2x 2 5 5 0 or x 1 3 5 0 Use the Zero-Product Property.

2x 5 5 or x 5 23 Solve for x.

x 552 or x 5 23

Check the solutions:

x 552: 2Q5

2R2

1 Q52R 0 15 x 5 23: 2(23)2 1 (23) 0 15

252 1

52 0 15 18 2 3 0 15

15 5 15 15 5 15

Both solutions check. Th e solutions are x 552 and x 5 23.

ExercisesSolve each equation by factoring. Check your answers.

1. x2 2 10x 1 16 5 0 2. x2 1 2x 5 63 3. x2 1 9x 5 22

4. x2 2 24x 1 144 5 0 5. 2x2 5 7x 1 4 6. 2x2 5 25x 1 12

7. x2 2 7x 5 212 8. 2x2 1 10x 5 0 9. x2 1 x 5 2

10. 3x2 2 5x 1 2 5 0 11. x2 5 25x 2 6 12. x2 1 x 5 20

4-5 ReteachingQuadratic Equations

2, 8

12

3, 4

23, 1

29, 7

2 12, 4

25, 0

23, 22

211, 2

24, 32

22, 1

25, 4


50

Name Class Date

Some quadratic equations are diffi cult or impossible to solve by factoring. You can use a graphing calculator to fi nd the points where the graph of a function intersects the x-axis. At these points f (x) 5 0, so x is a zero of the function.

Th e values r1 and r2 are the zeros of the function y 5 (x 2 r1)(x 2 r2). Th e graph of the function intersects the x-axis at x 5 r1, or (r1, 0), and x 5 r2, or (r2, 0).

Problem

What are the solutions of the quadratic equation 3x2 5 2x 1 7?

Step 1 Rewrite the equation in standard form, ax2 1 bx 1 c 5 0.3x2 2 2x 2 7 5 0

Step 2 Enter the equation as Y1 in your calculator.

Step 3 Graph Y1. Choose the standard window and see if the zeros of the function Y1 are visible on the screen. If they are not visible, zoom out and determine a better viewing window. In this case, the zeros are visible in the standard window.

Step 4 Use the ZERO option in the CALC feature. For the fi rst zero, choose bounds of 22 and 21 and a guess of 21.5. Th e screen display gives the fi rst zero as x 5 21.230139.

Similarly, the screen display gives the second zero as x 5 1.8968053.

Th e solutions to two decimal places are x 5 21.23 and x 5 1.90.

Exercises

Solve the equation by graphing. Give each answer to at most two decimal places.

13. x2 5 5 14. x2 5 5x 1 1

15. x2 1 7x 5 3 16. x2 1 x 5 5

17. x2 1 3x 1 1 5 0 18. x2 5 2x 1 4

19. 3x2 2 5x 1 9 5 8 20. 4 5 2x2 1 3x

21. x2 2 6x 5 27 22. 2x2 5 8x 1 8


Quadratic Equations

Plot1\Y153X2 2 2X 2 7\Y25\Y35\Y45\Y55\Y65\Y75

Plot2 Plot3

ZeroX 5 21.230139 Y 5 0

ZeroX 5 1.8968053 Y 5 0

22.24, 2.24

27.41, 0.41

22.62, –0.38

0.23, 1.43

1.59, 4.41

20.19, 5.19

22.79, 1.79

21.24, 3.24

22.35, 0.85

26.83, 21.17


59

Name Class Date

4-6 Reteaching Completing the Square

Completing a perfect square trinomial allows you to factor the completed trinomial as the square of a binomial.

Start with the expression x2 1 bx . Add Qb2R2

. Now the expression is x2 1 bx 1 Qb2R2

,

which can be factored into the square of a binomial: x2 1 bx 1 Qb2R2

5 Qx 1b2R

2.

To complete the square for an expression ax2 1 abx , fi rst factor out a. Th en fi nd the value that completes the square for the factored expression.

Problem

What value completes the square for 22x2 1 10x?

Exercises

What value completes the square for each expression?

1. x2 1 2x 2. x2 2 24x 3. x2 1 12x

4. x2 2 20x 5. x2 1 5x 6. x2 2 9x

7. 2x2 2 24x 8. 3x2 1 12x 9. 2x2 1 6x

10. 5x2 1 80x 11. 27x2 1 14x 12. 23x2 2 15x

Write

22x2 1 10x 5 22(x2 2 5x)

b2 5

252 5 2

52

22 cx2 2 5x 1 a25

2b2d 5 22ax2 2 5x 1

254 b

22Qx 252R

2

22Q254 R 5 2

252

Write the expression in the form a(x2 1 bx).

Find b2.

Add Qb2R2

to the inner expression to complete the square.


Find the value that completes the square.

Think

1

72

320

100254

814

2 754

144

12

27

36

29


60

Name Class Date

4-6 Reteaching (continued) Completing the Square

You can easily graph a quadratic function if you fi rst write it in vertex form. Complete the square to change a function in standard form into a function in vertex form.

Problem

What is y 5 x2 2 6x 1 14 in vertex form?

Exercises

Rewrite each equation in vertex form.

13. y 5 x2 1 4x 1 3 14. y 5 x2 2 6x 1 13

15. y 5 2x2 1 4x 2 10 16. y 5 x2 2 2x 2 3

17. y 5 x2 1 8x 1 13 18. y 5 2x2 2 6x 2 4

19. y 5 2x2 1 10x 2 18 20. y 5 x2 1 2x 2 8

21. y 5 2x2 1 4x 2 3 22. y 5 3x2 2 12x 1 8

Write

x2 2 6x

b2 5

262 5 23

x2 2 6x 1 (23)2 5 x2 2 6x 1 9

y 5 x2 2 6x 1 9 1 14 2 9

y 5 (x 2 3)2 1 14 2 9

y 5 (x 2 3)2 1 5

Write an expression using the terms that contain x.

Find b2.

Add Qb2R2

to the expression to complete the square.

Subtract 9 from the expression so that the equation is unchanged.


Add the remaining constant terms.

Think

2(x 2 2)2 2 6

(x 1 4)2 2 3

2(x 2 5)2 1 7

2(x 1 1)2 2 5

(x 1 2)2 2 1

(x 2 1)2 2 4

2(x 1 3)2 1 5

(x 1 1)2 2 9

3(x 2 2)2 2 4

(x 2 3)2 1 4


69

Name Class Date

4-7 ReteachingThe Quadratic Formula

You can solve some quadratic equations by factoring or completing the square.

You can solve any quadratic equation ax2 1 bx 1 c 5 0 by using the Quadratic Formula:

x 52b 4 "b2 2 4ac

2a

Notice the 4 symbol in the formula. Whenever b2 2 4ac is not zero, the Quadratic Formula will result in two solutions.

Problem

What are the solutions for 2x2 1 3x 5 4? Use the Quadratic Formula.

2x2 1 3x 2 4 5 0 Write the equation in standard form: ax2 1 bx 1 c 5 0

a 5 2; b 5 3; c 5 24 a is the coeffi cient of x2, b is the coeffi cient of x, c is the constant term.

x 52b 4 "b2 2 4ac

2a Write the Quadratic Formula.

52(3) 4 "(3)2 2 4(2)(24)

2(2) Substitute 2 for a, 3 for b, and −4 for c.

523 4 !41

4 Simplify.

523 1 !41

4 or 23 2 !414 Write the solutions separately.

Check your results on your calculator. Replace x in the original equation with 23 1 !41

4 and 23 2 !414 . Both values

for x give a result of 4. Th e solutions check.

Exercises

What are the solutions for each equation? Use the Quadratic Formula.

1. 2x2 1 7x 2 3 5 0 2. x2 1 6x 5 10

3. 2x2 5 4x 1 3 4. 4x2 1 81 5 36x

5. 2x2 1 1 5 5 2 7x 6. 6x2 2 10x 1 3 5 0

(23 1 V(41)) / 4 X .8507810594

2X2 1 3X 4

(23 2 V(41)) / 4 X 22.3507810592X2

1 3X 4

7 1 !372 or 7 2 !37

2

2 1 !102 or 2 2 !10

2

24 or 12

23 1 !19 or 23 2 !19

5 1 !76 or 5 2 !7

6

92


70

Name Class Date

There are three possible outcomes when you take the square root of a real number n:

n •. 0 S two real values (one positive and one negative)

5 0 S one real value (0)

, 0 S no real values

Now consider the quadratic formula: x 52b 4 "b2 2 4ac

2a . Th e value under the radical symbol determines the number of real solutions that exist for the equation ax2 1 bx 1 c 5 0:

b2 2 4ac •. 0 S two real solutions

5 0 S one real solution

, 0 S no real solutions

Problem

What is the number of real solutions of 23x2 1 7x 5 2?

23x2 1 7x 5 2

23x2 1 7x 2 2 5 0 Write in standard form.

a 5 23, b 5 7, c 5 22 Find the values of a, b, and c.

b2 2 4ac Write the discriminant.

(7)2 2 4(23)(22) Substitute for a, b, and c.

49 2 24 Simplify.

25

Th e discriminant, 25, is positive. Th e equation has two real roots.

Exercises

What is the value of the discriminant and what is the number of real solutions for each equation?

7. x2 1 x 2 42 5 0 8. 2x2 1 13x 2 40 5 0 9. x2 1 2x 1 5 5 0

10. x2 5 18x 2 81 11. 2x2 1 7x 1 44 5 0 12. 14x2 2 5x 1 25 5 0

13. 2x2 1 7 5 5x 14. 4x2 1 25x 5 21 15. x2 1 5 5 3x

16. 19x2 5 4x 2 36 17. 1

2x2 1 2x 1 3 5 0 18. 16x2 5 2x 1 18


The Quadratic Formula

Th e value under the radical, b2 2 4ac, is called the discriminant.

169; two

0; one

0; one

–31; none

225; two

961; two

–2; none

9; two –16; none

–11; none

16; two

0; one


79

Name Class Date

24 22 1 11i

3 1 12i

8 1 24i 99 1 45i

5 1 5i

6 1 3i 22i

3 1 10i

4-8 ReteachingComplex Numbers

• A complex number consists of a real part and an imaginary part. It is written in the form a 1 bi, where a and b are real numbers.

• i 5 !21 and i2 5 (!21)(!21) 5 21

• When adding or subtracting complex numbers, combine the real parts and then combine the imaginary parts.

• When multiplying complex numbers, use the Distributive Property or FOIL.

Problem

What is (3 2 i) 1 (2 1 3i)?

(3 2 i) 1 (2 1 3i)

5 3 2 i 1 2 1 3i Circle real parts. Put a square around imaginary parts.

5 (3 1 2) 1 (21 1 3)i Combine.

5 5 1 2i Simplify.

Problem

What is the product (7 2 3i)(24 1 9i)?

Use FOIL to multiply:

(7 2 3i)(24 1 9i) 5 7(24) 1 7(9i) 1 (23i)(24) 1 (23i)(9i)

5 228 1 63i 1 12i 2 27i2

5 228 1 75i 2 27i2

You can simplify the expression by substituting 21 for i2.

(7 2 3i)(24 1 9i) 5 228 1 75i 2 27(21)

5 21 1 75i

Exercises

Simplify each expression.

1. 2i 1 (24 2 2i) 2. (3 1 i)(2 1 i) 3. (4 1 3i)(1 1 2i)

4. 3i(1 2 2i) 5. 3i(4 2 i) 6. 3 2 (22 1 3i) 1 (25 1 i)

7. 4i(6 2 2i) 8. (5 1 6i) 1 (22 1 4i) 9. 9(11 1 5i)

First 5 7(24)

(7 2 3i)(24 1 9i)

Outer 5 7(9i)

Inner 5 (23i)(24)

Last 5 (23i)(9i)


80

Name Class Date

1 1 2i

3 1 i

265 1 3

5i

110 1 17

10i

3 2 5i

2 2 3i

917 2 15

17i

21113 1 10

13i

2i

25 1 2i

2 2 2i

23229 1 7

29i


Complex Numbers

• Th e complex conjugate of a complex number a 1 bi is the complex number a 2 bi.

• (a 1 bi)(a 2 bi) 5 a2 1 b2

• To divide complex numbers, use complex conjugates to simplify the denominator.

Problem

What is the quotient 4 1 5i2 2 i ?

4 1 5i2 2 i 5 The complex conjugate of 2 2 i is 2 1 i.

5 4 1 5i2 2 i ?

2 1 i2 1 i Multiply both numerator and denominator by 2 1 i.

5 8 1 4i 1 10i 1 5i2

(2 2 i)(2 1 i) Use FOIL to multiply the numerators.

5 8 1 4i 1 10i 1 5i2

22 1 12 Simplify the denominator. (a 1 bi)(a 2 bi) 5 a2 1 b2

58 1 14i 1 5(21)

4 1 1 Substitute 21 for i2.

5 3 1 14i5 Simplify.

535 1

145 i Write as a complex number a 1 bi.

Exercises

Find the complex conjugate of each complex number.

10. 1 2 2i 11. 3 1 5i 12. i

13. 3 2 i 14. 2 1 3i 15. 25 2 2i

Write each quotient as a complex number.

16. 3i1 2 2i 17. 6

3 1 5i 18. 2 1 2ii

19. 2 1 5i3 2 i 20. 24 2 i

2 1 3i 21. 6 1 i25 2 2i


89

Name Class Date

(21, 27), (1, 21)

(21, 22), (3, 10)

(1, 22), (4, 13)

(1, 3), (5, 21)

(25, 3), (1, 23)

(24, 25), (1, 0)

You used graphing and substitution to solve systems of linear equations. You can use these same methods to solve systems involving quadratic equations.

Problem

What is the solution of the system of equations? e y 5 x2 2 2x 2 8

y 5 2x 2 3

y 5 2x 2 3 Write one equation.

x2 2 2x 2 8 5 2x 2 3 Substitute x2 2 2x 2 8 for y in the linear equation.

x2 2 4x 2 5 5 0 Write in standard form.

(x 1 1)(x 2 5) 5 0 Factor the quadratic expression.

x 1 1 5 0 or x 2 5 5 0 Use the Zero-Product Property.

x 5 21 or x 5 5 Solve for x.

Because the solutions to the system of equations are ordered pairs of the form (x, y), solve for y by substituting each value of x into the linear equation. You can use either equation, but the linear equation is easier.

x 5 21: y 5 2x 2 3 5 2(21) 2 3 5 25 S (21, 25)x 5 5: y 5 2x 2 3 5 2(5) 2 3 5 7 S (5, 7)

Th e solutions are (21, 25) and (5, 7). Check these by graphing the system and identifying the points of intersection.

Exercises

Solve each system.

1. e y 5 x2 1 3x 2 5

y 5 3x 2 4 2. e y 5 2x2 1 5x 2 1

y 5 2x 1 4

3. e y 5 2x2 2 x 2 5

y 5 3x 1 1 4. e y 5 x2 1 3x 2 7

y 5 2x 2 2

5. e y 5 2x2 2 5x 1 1

y 5 5x 2 7 6. e y 5 2x2 2 2x 1 3

y 5 x 2 1

4-9 ReteachingQuadratic Systems


90

Name Class Date

�4�6�8

86

2

�4 4 6�8 8

y

x

�4�6�8

8

2

�4 4 6�8 8

y

x

�4�6�8

6

2

�4 4 6�8 8

y

x

�4�6�8

6y

x2

�4 2 6�8 8

To solve a system of linear inequalities, you graph each inequality and fi nd the region where the graphs overlap. You can also use this technique to solve a system of quadratic inequalities.

Problem

What is the solution of this system of inequalities? e y , 2x2 1 4x

y . x2 2 2x 2 8

Step 1 Graph the equation y 5 2x2 1 4x . Use a dashed boundary line because the points on the curve are not part of the solution. Choose a point on one side of the curve and check if it satisfi es the inequality.

y , 2x2 1 4x

0 ,?

2(2)2 1 4(2) Check the point (2, 0).0 , 4 The inequality is true.

Points below the curve satisfy the inequality, so shade that region.

Step 2 Graph the equation y 5 x2 2 2x 2 8. Use a dashed boundary line because the points on the curve are not part of the solution. Choose a point on one side of the curve and check if it satisfi es the inequality.

y . x2 2 2x 2 8

0 .?

(2)2 2 2(2) 2 8 Check the point (2, 0).0 . 28 The inequality is true.

Points above the curve satisfy the inequality, so shade that region.

Step 3 Th e solution to the system of both inequalities is the set of points satisfying both inequalities. In other words, the solution is the region where the graphs overlap. Th e region contains no boundary points.

Exercises

Solve each system by graphing.

7. e y , 2x2 1 6

y . x2 2 2 8. e y . x2 2 x 2 2

y , 2x2 2 x 1 6

9. e y , 2x2 2 2x 1 8

y . 2x2 1 4 10. e y . x2 2 6x

y , x2 2 6x 1 7


Quadratic Systems

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�2 2�4 4�6 6�8 8

y

x

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�2 2�4 4�6 6�8 8

y

x

�4�6

�2�4�6�8

�2 2�4 4�6 6�8 8

y

x

Quadratic Functions and Transformations€¦ · 4-1 Review Parent Quadratic Function The parent...

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