# What is the maximum value that the graph of f(x)= -x^2+8x+7?

Jul 25, 2016

I got $23$.

If you think about it, since ${x}^{2}$ has a minimum, $- {x}^{2} + b x + c$ has a maximum (that doesn't require closed bounds).

We should know that:

• The slope is $0$ when the slope changes sign.
• The slope changes sign when the graph changes direction.
• One way a graph changes direction is at a maximum (or minimum).
• Therefore, the derivative is $0$ at a maximum (or minimum).

So, just take the derivative, set it equal to $0$, find the value of $x$ (which corresponds to the maximum), and use $x$ to find $f \left(x\right)$.

$f ' \left(x\right) = - 2 x + 8$ (power rule)

$0 = - 2 x + 8$

$2 x = 8$

$\textcolor{b l u e}{x = 4}$

Therefore:

$f \left(4\right) = - {\left(4\right)}^{2} + 8 \left(4\right) + 7$

$= - 16 + 32 + 7$

$\implies \textcolor{b l u e}{y = 23}$

So your maximum value is $23$.

$y = - {x}^{2} + 8 x + 7$:

graph{-x^2 + 8x + 7 [-7.88, 12.12, 16.52, 26.52]}