Maximum value of a Quadratic Function I need help with the following problem:
Find the maximum value of the quadratic function $f(x)=ax^2+bx+c$ on the interval $[0,1],$ where $|f(-1)| \leq 1, |f(0)| \leq 1, |f(1)| \leq 1.$
Thanks for all tips!
 A: First, if, in $f(x)= ax^2+ bx+ c$, a is positive then the parabola opens upward so the maximum is at one of the endpoints.  f(0)= c and f(1)= a+ b+ c.  Which is larger?
If a is negative then then the maximum is in the interior where f'(x)= 2ax+ b= 0 so at x= -b/(2a).
A: The maximum (and minimum) of any quadratic function over any finite interval always occurs either at one of the endpoints of the interval or at the vertex.
Your question is unclear because it could be interpreted as finding the maximum for a fixed $a,b,c$ or it could be interpreted a the maximum over all $a,b,c$.
For the former interpretation:
You have to break it down into separate cases defined by conditions on $a,b,c$. The maximum will be either $f(0)=c$, $f(1)=a+b+c$, or $f(-b/(2a))=c-b^2/(4a)$. You just have to figure out which case is which.  If $0<-b/(2a)<1$ and $a<0$, the maximum will be $c-b^2/(4a)$. Otherwise, the maximum will be $\max \{c,a+b+c \}$.
For the latter interpretation:
The maximum either occurs at $0$, $1$ or at the vertex if the x-coordinate of the vertex is in $[0,1]$.
It is already given that if the maximum is at $0$ or $1$, then the maximum is at most 1. So, find the maximum possible value assuming the x-coordinate of the vertex is in $[0,1]$.
There are several conditions:
$-1 \le a-b+c \le 1$
$-1 \le c \le 1$
$-1 \le a+b+c \le 1$
$0 \le -b/(2a) \le 1$
It turns out the maximum is when $a=-1$, $b=c=1$ and in that scenario, the vertex is $(1/2, 5/4)$.
So the maximum is $5/4$.
