What is the vertex of the function f(x) = x2 + 12x?

2 answers:

Volgvan3 years ago

8 0

Answer:

A: (-6, -36)

Step-by-step explanation:

This should be the right answer on edge,

please correct me if I'm wrong.

5 0

Answer:

6, - 36 )

Step-by-step explanation:

Given a parabola in standard form

y = ax² + bx + c ( a ≠ 0 )

Then the x- coordinate of the vertex is

x = - $\frac{b}{2a}$

f(x) = x² + 12x ← is in standard form

with a = 1, b = 12, c = 0 , then

x = - $\frac{12}{2}$ = - 6

Substitute x = 6 into f(x) for corresponding value of y

f(6) = 6² + 12(- 6) = 36 - 72 = - 36

vertex = (6, - 36 )

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ch4aika [34]

Answer:

The number that belongs in the green box is equal to 909.

General Formulas and Concepts:
Algebra I

Equality Properties

Trigonometry

[Right Triangles Only] Pythagorean Theorem:
$\displaystyle a^2 + b^2 = c^2$

Step-by-step explanation:

Step 1: Define

Identify given variables.

a = 30

b = 3

c = x

Step 2: Find x

Let's solve for the general equation that allows us to find the hypotenuse:

[Pythagorean Theorem] Square root both sides [Equality Property]:
$\displaystyle \begin{aligned}a^2 + b^2 = c^2 \rightarrow c = \sqrt{a^2 + b^2}\end{aligned}$

Now that we have the formula to solve for the hypotenuse, let's figure out what x is equal to:

[Equation] Substitute in variables:
$\displaystyle \begin{aligned}c & = \sqrt{a^2 + b^2} \\x & = \sqrt{30^2 + 3^2}\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}c & = \sqrt{a^2 + b^2} \\x & = \sqrt{30^2 + 3^2} \\& = \boxed{ \sqrt{909} } \\\end{aligned}$