Explanation
The problem is asking for two function equations of the "quadratic" graph, the factored form equation and the vertex form equation.
Vertex form equation) This functional equation has always the form
[tex]f(x)=a\cdot(x-h)^2+k[/tex]
It is called vertex form for (h,k) is the vertex of the parabola. Intuitively, the vertex of a parabola is its "starting" point, the point from where the parabola opens its "arms".
In this case, as you can see in the picture above, the vertex is
[tex](h,k)=(2,-8)[/tex]
Then,
[tex]h=2,k=-8[/tex]
Thus our equation becomes
[tex]f(x)=a\cdot(x-2)^2-8[/tex]
Our task now is to find the constant a. Note that the origin (0,0) is part of our function, that is
[tex]f(0)=0[/tex]
This (equality) is the same as
[tex]0=f(0)=a\cdot(0-2)^2-8[/tex]
Then
[tex]\begin{gathered} 0=a\cdot(-2)^2-8, \\ 8=a\cdot(-2)^2, \\ 8=a\cdot4,\text{ for }(-2)^2=4 \\ a=\frac{8}{4}=2 \end{gathered}[/tex]
Factored form equation) This equation is easier. It has the form
[tex]f(x)=b(x-r_1)(x-r_2),[/tex]
where r_1 and r_2 denote the roots of the parabola. They are just the x-intercepts of the function.
In this case, as you can see in the picture above,
[tex]r_1=0,r_2=4[/tex]
Then, our equation becomes
[tex]f(x)=b\cdot x\cdot(x-4)[/tex]
For the vertex of the parabola is clearly a point of the function, we have
[tex]f(2)=-8[/tex]
So
[tex]-8=f(2)=b\cdot2\cdot(2-4)[/tex]
Let's solve this for b:
[tex]\begin{gathered} -8=f(2)=b\cdot2\cdot(2-4), \\ -8=b\cdot2\cdot(-2), \\ -8=b\cdot(-4), \\ b=\frac{-8}{-4}, \\ b=2 \end{gathered}[/tex]Answer
The vertex form equation of f(x) is
[tex]f(x)={{\textcolor{red}{2}}\cdot(x-{\textcolor{red}{2}})}^2\textcolor{red}{-8}[/tex]
(Be careful! In the third box, you must put -8 not just 8. In the second one, you must put just 2 without the minus sign)
The factored form equation of f(x) is
[tex]f(x)=\textcolor{red}{2}\cdot x\cdot(x-\textcolor{red}{4})[/tex]
