Find the equation, f(x) = a(x-h)2 + k, for a parabola that passes through the point (0, 0) and has (-3, -6) as its vertex. What is the standard form of the equation?

A) The vertex form of the equation is f(x) = −2/3(x + 3)2 - 6. The standard form of the equation is f(x) = 2/3x2 + 4x.

B) The vertex form of the equation is f(x) =2/3(x + 3)2 - 6. The standard form of the equation is f(x) = −2/3x2 - 4x.

C) The vertex form of the equation is f(x) =−2/3(x - 3)2 - 6. The standard form of the equation is f(x) = 2/3x2 - 4x.

D) The vertex form of the equation is f(x) = 2/3(x + 3)2 - 6. The standard form of the equation is f(x) = 2/3x2 + 4x.

The answer is D because the value of a (2/3) is positive, and only D has it positive in both forms.

Step-by-step explanation:

The vertex form of the equation for a parabola is a(x – h)² + k, and the vertex is (h,k).

The standard form of the parabola equation is  ax² + bx + c.

From your given information, the vertex is (-3, -6) so by plugging that into the equation, we get a[x – (-3)]² + (-6) = a(x + 3)² - 6

so f(x) = a(x +3)² - 6 and we put in the point (0,0) we get

0 = a(0 +3)² - 6 and solve for a.

0 = a(3)² - 6

 0 = 9a -6  (add 6 to both sides to get a by itself)

+6         +6

6   = 9a  (divide both sides by 9 to get a by itself)

9      9

6/9 = a  (we simplify 6/9 by reducing both numbers by 3 to get 2/3)

a = 2/3  therefore our vertex form is now f(x) = (2/3)(x +3)² - 6.

We have to expand our equation to find the standard form.

(x +3)² = (x+3)(x+3) = x² + 3x + 3x +9 = x² +6x+ 9 (plug it back in)

(2/3)(x² +6x+ 9) - 6

2/3x² + (2/3)6x+ (2/3)9  - 6

2/3x² +4x+ 6   - 6

f(x) = 2/3x² + 4x is standard form; vertex form is now f(x) = (2/3)(x +3)² - 6

sorry, i can't understand your question. will later if you describe more clearly!

: )

answer:

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