Is the relation a function? {(-6, 3), (-1,0), (2, 4), (-1, -7), (5, 2)}
1 answer:
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Answer:
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
<u>Algebra I</u>
- Standard Form: ax² + bx + c = 0
- Quadratic Formula:
<u>Algebra II</u>
- Imaginary roots: √-1 = i
Step-by-step explanation:
<u>Step 1: Define function</u>
f(x) = -x² - 6x - 14
<u>Step 2: Set up</u>
- Set equation equal to 0: -x² - 6x - 14 = 0
- Factor out -1: -(x² + 6x + 14) = 0
- Divide both sides by -1: x² + 6x + 14 = 0
<u>Step 3: Define variables</u>
a = 1
b = 6
c = 14
<u>Step 4: Find roots</u>
- Substitute:
- Exponents:
- Multiply:
- Subtract:
- Factor:
- Simplify:
- Factor:
- Divide:
Answer:
Step-by-step explanation:
y = -x + 7 -------------(II)
Plugin y = -x + 7 in equation (I)
2(-x + 7) =x² - 6x + 9
-2x + 14 =x ² - 6x + 9
x² - 6x + 9 = -2x + 14
x² - 6x + 9 + 2x - 14 = 0
x² - 4x - 5 = 0
x² -5x + x - 5 = 0
x(x - 5) + 1(x - 5) = 0
(x -5)(x + 1) = 0
x- 5 = 0 ; x + 1 = 0
x = 5 ; x = -1
When x = 5 ⇒ y = -5 + 7 = 2
When x = -1 ⇒ y = -(-1) + 7 = 1 + 7 = 8
Solution (5 , 2) & (-1 , 8)