Question 1116065
The revenue function is
{{{r(x)=-500(x-2.5)^2+"28,025"}}} .
 
When you do not remember "formulas"
and other calculation tricks and problem-solving recipes taught in class,
using your own brain may work, if a little slower.
 
You do not need to have been awake during the classes about quadratic functions to know that
{{{(x-2.5)^2>=0}}} no matter what real number value {{{(x-2.5)}}} and {{{x}}} take;
{{{-500(x-2.5)^2<=0}}} no matter what, and as a consequence
{{{r(x)=-500(x-2.5)^2+"28,025"<=0+"28,025"="28,025"}}} no matter what.
 
You should also realize that the only way to have {{{r(x)="28,025"}}}
is to have {{{-500(x-2.5)^2=0}}} ,
which means {{{(x-2.5)^2=0}}} ,
{{{x-2.5=0}}} , and {{{highlight(x=2.5)}}} .
 
So, if (and only if)
the movie theater lowers ticket prices by {{{highlight(2.5)}}} dollars,
they can expect the maximum predicted revenue of {{{"$28,025"}}} .
 
NOTE: Quadratic functions graph as parabolas,
with a maximum (if the coefficient of the squared term is negative),
or a minimum (if that coefficient is positive).
They can be found in the form
{{{f(x)=a(x-h)^2=k}}} or {{{f(x)=ax^2+bx+c}}} ,
Where a,b,c,h, and k are constants,
but it must be {{{a<>0}}} (or it is not a quadratic function).
You can also find them in any other equivalent form that your teacher could choose.
It will be up to you to recognize what you are dealing with.
If your teacher is merciful, the function will look like {{{f(x)=a(x-h)^2=k}}} ,
and as explained at the top {{{x=h}}} will be
the {{{x}}} value where the maximum or minimum happens,
and that maximum or minimum value of the function will be {{{k}}} .
If the quadratic function appears in the form {{{f(x)=ax^2+bx+c}}} ,
you could transform it to the equivalent, but more useful form,
or you could remember that the maximum or minimum happens for'{{{x=(-b)/"2 a"}}} .