\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

Asked Apr 1, 2011 by Hayden (270 points) edited Jun 15, 2013 by Community

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}

Answered Apr 1, 2011 by Matt Sullivan (1,160 points) edited Jun 15, 2013 by Community

Answered Apr 1, 2011 by Matt Sullivan (1,160 points)

## Maths Methods Quadratics Question

Answered## 2 Answers

So: 3=-2p/-2

p=3

Use the same method as before to find q

Answered

Apr 1, 2011by Matt Sullivan (1,160 points)Notice: Undefined index: title in/home3/wmroi/public_html/merspi.com.au/qa-theme/NewMerspi/qa-theme.phpon line1251