Find b and c so that has vertex .
I really dont even know where to start
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
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Vertex of a quadratic curve
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
$\displaystyle \left\{\begin{array}{ll}\displaystyle\frac{b}{10}= 5\\\displaystyle\frac{b^2+20c}{20}=-2\end{array}\right.$