# Trouble finding b and c. I know the vertex.

• September 8th 2009, 10:55 AM
dtownangel
Trouble finding b and c. I know the vertex.
Find b and c so that http://math.mscd.edu/webwork2_files/...b9c1ea4471.png has vertex http://math.mscd.edu/webwork2_files/...37aaf10c91.png.
I really dont even know where to start
• September 8th 2009, 11:17 AM
Amer
Quote:

Originally Posted by dtownangel
Find b and c so that http://math.mscd.edu/webwork2_files/...b9c1ea4471.png has vertex http://math.mscd.edu/webwork2_files/...37aaf10c91.png.
I really dont even know where to start

the vertex of quadratics function given by $\frac{-b}{2a}$

here we have a=-5 and b=b and c=c so

you want the vertex have the x-coordinate equal 5 so

$\frac{-b}{2(-5)} = 5 \rightarrow b=50$

we have now $y=-5x^2+50x+c$ when x=5 y=-2

$-2=-5(25)+50(5)+c$ find c value

If $y=ax^2+bx+c$ then the vertex has the coordinates $\left(-\frac{b}{2a},-\frac{\Delta}{4a}\right)$, where $\Delta=b^2-4ac$.
In this case $a=-5, \ \Delta=b^2+20c$.
$\left\{\begin{array}{ll}\displaystyle\frac{b}{10}= 5\\\displaystyle\frac{b^2+20c}{20}=-2\end{array}\right.$