How can I prove why I chose the point to lie on a parabola? I have a problem: Let $A$ and $B$ be two points that lie on the parabola $y=x^2 + 3x - 4$, where $x \in [-3,3]$. Find the greatest value of the segment $AB$.
Here is what I tried: Suppose $B=(3,14)$ and $A=(a, a^2 +3a - 4)$. We have 
$$AB^2 = a^4+6 a^3+46 a^2+102 a+333.$$
From here, I have maximum of $AB^2$ is appr. 1296.
How can I prove why I chose the point $B$ to lie on the parabola like that?
 A: Write it as $y=f(x)=\left(x+3/2\right)^2 - \frac{25}{4}$.
This means that $f(x)=f(-3-x)$. We have to take care with this symmetry, because $x$ and $-3-x$ are both in $[-3,3]$ if any only if $x\in[-3,0]$.
Assume $-3\leq a\leq b\leq 3$.
Now, if $b\leq\frac{-3}{2}$, then the distance between $(a,f(a))$ and $(b,f(b))$ is the same as the distance between $(a',f(a'))$ and $(b',f(b'))$, with $a'=-3 -b, b'=-3-a$, and we still have $a',b'\in[-3,3]$ and $a'<b'$, but with $b'\geq \frac{-3}{2}$.
So we can assume that $b\geq\frac{-3}{2}$.
If $f(a)> f(b)$, then show that you can start with $b'=-3-a, a'=-3-b$. This is the tricky one. if $f(a)>f(b)$ and $a<b$ and $b\geq -3/2$, then $a$ must be further from $-3/2$ than $b$ is. But $a$ must be at most $3/2$ less than $-3/2$, so $-3-a$ is still in $[-3,3]$, and the same is true for $-3-b$.
This means we can assume $f(a)\leq f(b)$. But then, since $b\geq-3/2$, increasing $b$ increases both $|f(b)-f(a)|$ and $|b-a|$, and hence the distance, so you can assume $b=3$.

This might seem complicated, but "visually" it is fairly simple. 
If you draw the graph of the function, then it is symmetric in the interval $[-3,0]$ with center $-3/2$. 
So you can assume both points are not in the $[-3,3/2]$ range. Likewise, if both points are in the $[-3,0]$ range, with $a\leq -3/2 \leq b\leq 0$, you can apply the symmetry to assume that $f(a)\leq f(b)$. 
Also, if $b>0$ and$-3\leq a\leq b$ then $f(a)\leq f(b)$, because $b$ is further from $-3/2$ than $a$ is. 
But if $f(a)\leq f(b)$ and $a<b$ and $b\geq -3/2$, then increasing $b$ increases the distance, so the maximum distance must occur when $b=3$.

By the way, the same argument also means that $a\leq -3/2$ - because the distance is clearly increasing as $a$ moves from $3$ to $-3/2$.
