Find b and c so that the parabola has vertex .
What do I do here?
The method I was thinking about is the following.
Given is the parabola: $\displaystyle y=-4x^2+bx+c$ and the vertex point with coordinates $\displaystyle (9,10)$
In general we know we can directly read the coordinates of the vertex of a parabola if we convert $\displaystyle y$ to:
$\displaystyle y=a(x-h)^2+q$ where $\displaystyle (h,q)$ are the coordinates of the parabola.
First, divide every term by $\displaystyle -4$ so:
$\displaystyle \frac{y}{-4}=x^2-\frac{b}{4}x-\frac{c}{4}$
Use completing the square in the RHS:
$\displaystyle \frac{y}{-4}=\left(x-\frac{b}{8}\right)^2-\frac{b^2}{64}-\frac{c}{4}$
$\displaystyle \Leftrightarrow y=-4\left(x-\frac{b}{8}\right)^2+\frac{b^2}{16}+c$
That means:
(1)$\displaystyle \frac{b}{8}=9 \Leftrightarrow b=72$ (1)
(2)$\displaystyle \frac{b^2}{16}+c=10$ (2)
Substituting (1) in (2) gives:
$\displaystyle 324+c=10 \Leftrightarrow c=-314$
Therefore the parabola with vertex $\displaystyle (9,10)$ is:
$\displaystyle y=-4x^2+72x-314$
But as you see, the method sudharaka wants to use is easier.