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:
$$f(x) = ax^2 + bx + c$$
 A: Hint: solve $\,f(2)=0\,, \;f'(2)=0\,, \;f(-1)=-6\,$ for $\,a,b,c\,$.
A: Here is another approach. Since the parabola is symmetric about the axis of symmetry, we know that $(5, -6)$ is another point on the parabola. Since $f(5)=-6=f(-1)$ and $f(2)=0$, it follows that
\begin{align}
4a + 2b + c &= 0\\
a-b+c &= -6\\
25a+5b+c&=0
\end{align}
which you can solve (for example by Gaussian elimination) to get $a=-2/3, b=8/3$ and $c=-8/3$.
A: *

*Start by using the "shifted" or "vertex" form $f(x) = a(x - h)^2 + k$; since you know the vertex, you know everything but $a$. 

*You can find $a$ since you know $f(-1) = -6$; set up and solve that equation for $a$. 
Then just multiply everything out to get the form you're after. (Also the title is a bit misleading; there's only a "highest point" on a parabola if it opens downward, and in this case, the highest point is the vertex.)
A: $$(x_v,y_v)=\left(-\frac{b}{2a},-\frac{\Delta}{4a}\right)=(2,0)$$
$b=-4a$ and $\Delta=0 \Rightarrow b^2=4ac$. 
$$16a^2=4ac \Rightarrow 4a(4a-c)=0$$
But, $a\ne 0$ so $c=4a$.
$$f(x)=a(x^2-4x+4)=a(x-2)^2$$
But $f(-1)=-6$ so $a=-2/3$
$$f(x)=-\frac{2}{3}(x-2)^2$$
A: 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.
A: Vertex form of a quadratic is: $f(x)=a(x-h)^2+k$, where $(h,k)$ is the vertex.  So, in our case the vertex is $(2,0)$ so $h=2$ and $k=0$.  So we have, $f(x)=a(x-2)^2$.  Now, find $a$ by using the point $(-1,-6$).  So we have, $-6=a(-1-2)^2$ which gives us $-6=9a$ so $a=-\frac{2}{3}$.  So our equation is $f(x)=-\frac{2}{3}(x-2)^2$.  Now, multiply this out to get $$f(x)=-\frac{2}{3}x^2+\frac{8}{3}x-\frac{8}{3}$$
