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

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- Sep 8th 2009, 10:55 AMdtownangelTrouble 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 - Sep 8th 2009, 11:17 AMAmer
the vertex of quadratics function given by $\displaystyle \frac{-b}{2a}$

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

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

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

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

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

for more information see this link

Vertex of a quadratic curve - Sep 8th 2009, 11:22 AMred_dog
If $\displaystyle y=ax^2+bx+c$ then the vertex has the coordinates $\displaystyle \left(-\frac{b}{2a},-\frac{\Delta}{4a}\right)$, where $\displaystyle \Delta=b^2-4ac$.

In this case $\displaystyle a=-5, \ \Delta=b^2+20c$.

Then you have to solve the system

$\displaystyle \left\{\begin{array}{ll}\displaystyle\frac{b}{10}= 5\\\displaystyle\frac{b^2+20c}{20}=-2\end{array}\right.$