<s>[INST] How to find the equation of a parabola when only given the x-intercepts and the Axis of symmetry? i have been given the following problem. 

A tennis ball is lobbed from ground level and must cover a horizontal
  distance of 22m if it is to land just inside the opposite end of the 
  court. If the opponent is standing 4m from the baseline and he can hit 
  any ball less than 3m high, what is the lowest maximum height the lob
  must reach to win the point.

The answer is 5 and 1/24 Metres (back of the textbook)
I have no idea how to approach this problem but my attempt to
approach this problem is by finding the X-intercepts which are 0, 
22. Thus, finding the Axis of symmetry that is 22/2 = 11. That is 
all i've done so far. 
Explanation will be extremely helpful :)
 [/INST] The general equation of a parabola is
$$
f(x) = a x^2 + bx + c
$$
it has three parameters.
We know $f(11) = 0$, so we have
$$
f(11) = 0
$$
we also know
$$
3 = f(11-4) = f(7)
$$
further we lobbed from ground level, such that we are 22m away, so this means
$$
0 = f(-11) = a 11^2 - 11 b + c
$$
This gives the linear system in the unknowns $a,b,c$:
$$
\left[
\begin{array}{rrr|r}
121 & 11 & 1 & 0 \\
49 & 7 & 1 & 3 \\
121 & -11 & 1 & 0
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
242 & 0 & 2 & 0 \\
49 & 7 & 1 & 3 \\
121 & -11 & 1 & 0
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 0 & 1/121 & 0 \\
49 & 7 & 1 & 3 \\
0 & -11 & 0 & 0
\end{array}
\right]
\to
\\
\left[
\begin{array}{rrr|r}
1 & 0 & 1/121 & 0 \\
0 & 1 & 0 & 0 \\
49 & 0 & 1 & 3 \\
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 0 & 1/121 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1-49/121 & 3 \\
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 0 & 1/121 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 72/121 & 3 \\
\end{array}
\right]
\to 
\\
\left[
\begin{array}{rrr|r}
1 & 0 & 0 & -1/24 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 121/24 \\
\end{array}
\right]
$$
The highest point is achieved in the middle, having height
$$
f(0) = c = 121/24
$$

</s>