<s>[INST] How can I find the quadratic equation from highest point on parabola? If I have a parabola, where the vertex is in $P=(2,0)$, and a point on the parabola is $Q=(-1,-6)$, how can I find the quadratic equation of the form:
$$f(x) = ax^2 + bx + c$$
 [/INST] The vertex is the
extreme point,
at which $f'(x) = 0$.
Since
$f'(x) = 2ax+b$,
this is zero when
$x = -\frac{b}{2a}
$.
At this point
$\begin{array}\\
f(x)
&=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\
&=\frac{ab^2}{4a^2}-\frac{b^2}{2a}+c\\
&=\frac{b^2}{4a}-\frac{b^2}{2a}+c\\
&=-\frac{b^2}{4a}+c\\
\end{array}
$
Since this point is
$(2, 0)$,
$-\frac{b}{2a}=2$
and
$-\frac{b^2}{4a}+c = 0$,
or
$b = -4a$
and
$c = \frac{b^2}{4a}
=\frac{16a^2}{4a}
=4a
$.
Since the parabola
also passes through
$(-1, -6)$,
we get
$-6
= a(-1)^2+b(-1)+c
= a-b+c
=a-(-4a)+4a
=9a
$
or
$a=-\frac{6}{9}=-\frac{2}{3}$.
$b$ and $c$ follow.
</s>