Question

**PLEASE HELP**

Find the equation, f(x)=a(x-h)^2+k, for a parabola that passes through the point (0,0) and has a (-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)=2/3x^2+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)+-2/3x^2-4x.

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)=2/3x^2-4x.

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

Answer 1

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

Explanation:

Let
`A(x,y) = (0,0)` and
`V(x,y) = (-3, -6)` (vertex).

If we know that
`x = 0`,
`y = 0`,
`h = -3` and
`k = -6`, then the standard equation results in:

`a\cdot (0+3)^(2)-6 = 0`

And the value of
`a` is:

`9\cdot a - 6 = 0`

`a = (2)/(3)`

And the vertex form of the equation is
`f(x) = (2)/(3)\cdot (x+3)^(2)-6`.

Lastly, the standard form of the equation is found by algebraic means:

`f(x) = (2)/(3)\cdot (x^(2)+6\cdot x +9)-6`

`f(x) = (2)/(3)\cdot x^(2)+4\cdot x`

The standard form of the equation is
`f(x) = (2)/(3)\cdot x^(2)+4\cdot x`.

In consequence, the right answer is D.