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\begin{equation}y=-x^2+2px+q\end{equation} has a maximum value of 5 when x = 3.

Find the values of p and q


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2 Answers

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First of all:

\begin{equation}
dy/dx=-2x+2p
\end{equation}

We know that dy/dx=0 for x=3 as this means the gradient is 0 (turning point)

So:

\begin{equation}
0=-2 * 3 + 2p
\end{equation}

\begin{equation}
0=-6+2p
\end{equation}

\begin{equation}
2p=6
\end{equation}

\begin{equation}
p=3
\end{equation}

Now we know the graph goes through the point (3,5) and p=3 so the equation becomes:

\begin{equation}
5 = -(3^2) + 2 * 3 * 3 + q
\end{equation}

\begin{equation}
5=-9+18+q
\end{equation}

\begin{equation}
q=-4
\end{equation}


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You could also do this by using x=-b/2a where x is the x value of the turning point

So: 3=-2p/-2

p=3

Use the same method as before to find q

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