Question 753632
{{{f(x)= (-4/7)(x-4)^2+8}}} is a quadratic function, so its graph is a parabola.
{{{f(x)= (-4/7)(x-4)^2+8}}} is the equation of that parabola in vertex form.
The vertex form is the form that makes it easiest to find the vertex.
When {{{x=4}}} {{{x-4=0}}} and {{{f(x)= (-4/7)(x-4)^2+8=(-4/7)0^2+8=8}}}
For all other values of {{{x}}},
{{{(x-4)^2>0}}}, {{{(-4/7)(x-4)^2<0}}}, and {{{f(x)= (-4/7)(x-4)^2+8<=8}}}
No matter what value {{{x}}} takes, {{{f(x)<=8}}}
and {{{f(x)=8}}} only when {{{x=4}}}.
The function has a maximum at {{{x=4}}},
and the value of that maximum is {{{f(4)=8}}}.
That corresponds to the vertex of the parabola, the point (4,8).
The function does not have a minimum; it can take any negative value you can think of.
The graph looks like this:
{{{graph(300,300,-6,14,-10,10,(-4/7)(x-4)^2+8)}}}
Parabolas can look like this {{{graph(100,100,1,7,3,9,(-4/7)(x-4)^2+8)}}} or like this {{{graph(100,100,1,7,-9,-3,4/7*(x-4)^2-8)}}}
They can have a maximum or a minimum, but not both, and whichever they have happens at the vertex.