The graph of the function f(x) = ax^2+bx+c contains the points (p - q, q), (p, -2q), and (p+q, q) with q =/= 0. Determine a, b, and c in terms of p's and q's.
I have no idea how to do this, please help! An explanation of how to get to the answer would be nice. Thanks in advance!
A point $(x,y)$ is "on" the graph of $f(x)$ if $y = f(x)$. So if $(p - q, q)$ is on the graph of $f(x) = ax^2 + bx + c$, then it must be that $q = a(p - q)^2 + b(p - q) + c$. You can get two more equations like this, using the other two points. Now you have three equations, with only three unknowns - remember, $p$ and $q$ don't count as "unknown", because you're allowed to leave them in your answer. So you can use these three equations to solve for $a$, $b$, and $c$.
Since $p$ is the midpoint of the line segment $\overline{(p-q)(p+q)}$ , the apex of the graph is $(p;-2q)$. Thus
$f(x)=a(x-p)^2-2q$.
From $q=f(p+q)$ we easily get $a=\frac{3}{q}$ and then $b=-\frac{6p}{q}$ and $c=\frac{3p^2}{q}-2q$
