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

Answer:

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

Step-by-step 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:

$9\cdot a - 6 = 0$

$a = \frac{2}{3}$

And the vertex form of the equation is $f(x) = \frac{2}{3}\cdot (x+3)^{2}-6$.

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

$f(x) = \frac{2}{3}\cdot (x^{2}+6\cdot x +9)-6$

$f(x) = \frac{2}{3}\cdot x^{2}+4\cdot x$

The standard form of the equation is $f(x) = \frac{2}{3}\cdot x^{2}+4\cdot x$.  

In consequence, the right answer is D.