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
(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
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 = 72x² + 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 = 0x² - 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
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
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 - 6x² - x - 6
The factors are: (x - 3) (x + 2)
x - 3 = 0 & x + 2 = 0
Roots are: x = 3 & x = -2
X-Intercepts
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
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
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
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
-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