Quadratic Functions

Lauren Sullivan

I. FINDING SOLUTIONS

Using Factoring

Find the factors and the x-intercepts of: x² - 7x + 10 = 0

x² - 7x + 10 = 0

(x - 2) (x - 5) = 0

x = 2 & x = 5

Using Square Root Method - Real Solution

Find the solutions by using the square root method: (x + 2)² - 16 = 0

(x + 2)² - 16 = 0

+16 +16

(x + 2)² = 16

√(x + 2)² = √16

x + 2 = +- 4

- 2 - 2

x = -2 +- 4

Using Square Root Method - Imaginary Solution

Find the solutions using the square root method: 2x² + 16x + 72 = 0

2x² + 16x + 72 = 0

/2 /2 /2

x² + 8x + 36 = 0

-36 -36

x² + 8x ______ = -36 (8 / 2)² = 16

√x² + 8x + √16 = -36

+16 +16

√(x + 4)² = √-20

x = -4 +- 2i√5

Using the Quadratic Formula - Real Solution

ax² + bx + c = 0 x = -b +- √b² - 4ac / 2a

Find the roots by using the quadratic formula: 2x² + 6x = 7

2x² + 6x = 7

-7 -7

2x² + 6x - 7 = 0

x = -6 +- √36 - (4)(2)(-7) / 4

x = -6 +- √36 + 56 / 4

x = -6 +- √92 / 4

x = -6 +- 2√23 / 4

x = -3 +- √23 / 2

Using the Quadratic Formula - Imaginary Solution

ax² + bx + c = 0 x = -b +- √b² - 4ac / 2a

Find the roots by using the quadratic formula: x² - 6x + 12 = 0

x² - 6x + 12 = 0

x = 6 +- √36 - (4)(1)(12) / 2

x = 6 +- √36 - 48 / 2

x = 6 +- √-12 / 2

x = 6 +- 2i√3 / 2

x = 3 +- i√3

By Completing the Square - Real Solution

Find the zeros by completing the square: x² + 6x - 27 = 0

x² + 6x - 27 = 0

+27 +27

x² + 6x ____ = 27 (6 / 2)² = 9

x² + 6x + 9 = 27

+9 +9

√(x + 3)² = √36

x + 3 = 6

-3 -3

x = -3 + - 6

By Completing the Square - Imaginary Solution

Find the zeros by completing the square: x² + 6x + 14 = 0

x² + 6x + 14 = 0

-14 -14

x² + 6x _____ = -14 (6 / 2)² = 9

√x² + 6x + √9 = -14

+9 +9

√(x + 3)² = √-5

x = -3 +- i√5

II. DISCRIMINANT

One Real Solution

ax² + bx + c = 0 b² - 4ac

Find the value of the discriminant and tell the number and nature of the solution:

x² - 8x + 16 = 0

8² - (4)(1)(16) = ?

64 - 64 = 0

1 Real Solution

Two Real Solutions

ax² + bx + c = 0 b² - 4ac

Find the value of the discriminant and tell the number and nature of the solution:

4x² - 12x - 10 = 0

12² - (4)(4)(-10) = ?

144 - (-160) = 304

2 Real Solutions

Two Imaginary Solutions

ax² + bx + c = 0 b² - 4ac

Find the value of the discriminant and tell the number and nature of the solution:

2x² - 8x + 16 = 0

8² - (4)(2)(16) = ?

64 - 128 = -64

2 Imaginary Solutions

III. TYPES OF ANSWERS

Roots

To find the roots we set each factor equal to 0 and solve for x:

Example: x² - x - 6

x² - x - 6

The factors are: (x - 3) (x + 2)

x - 3 = 0 & x + 2 = 0

Roots are: x = 3 & x = -2

X-Intercepts

To find the x-intercepts are the roots, zeros, and solutions.

Example: f(x) = x² - x - 6

f(x) = x² - x - 6

(graph the function and find the x-intercpets)

X-Intercepts are: x= 3 & x= -2

Zeros

To find the zeros we set each factor equal to 0 and solve for x:

Example: x² - x - 6

x² - x - 6


The factors are: (x - 3) (x + 2)

x - 3 = 0 & x + 2 = 0

Roots are: x = 3 & x = -2


Solutions

We make the polynomial expression into an equation solve for x:

Example: x² - x - 6 = 0

x² - x - 6 = 0

The factors are: (x - 3) (x + 2) = 0

x - 3 = 0 & x + 2 = 0

Solutions are: x = 3 & x= -2

IV. VERTEX FORM OF A QUADRATIC FUNCTION

With no Number in Front of the Term

Convert the standard form of the quadratic to vertex form: f(x) = x² - 6x - 2

x² - 6x - 2 = 0

+2 +2

x² - 6x _____ = 2 (6 / 2)² = 9

√x² - 6x + √9 = 2

+9 +9

(x - 3)² = 11

-11 -11

f(x) = (x - 3)² - 11

With a Number in Front of the Term

Convert the standard form of the quadratic to vertex form: -2x² - 4x + 6

-2x² - 4x + 6 = 0

/-2 /-2 /-2

x² + 2x - 3 = 0

+3 +3

x² + 2x _____ = 3 (2 / 2)² = 1

x² + 2x + 1 = 3

+1 +1

√x² + 2x + √1 = 4

(x + 1)² - 4

f(x) = -2(x + 1)² - 4

V. TRANSFORMATION

Vertical Shift

Example: x² + 6
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Horizontal Shift

Example: (x - 2)²
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Vertical Stretch

Example: -2 (x + 1)² - 4
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Horizontal Stretch

Example: - 2/3 (x - 1)² + 8


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Reflections over the X-Axis

Example: - (x - 3)² - 11
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