Calculus and Its Applications, Chapter 1, Review Exercises, Section Review Exercises, Problem 34

Below is the graph of the function $y = g(x)$. Determine the points on the graph of $y = -x^2 + 8x -11$
at which the tangent line is horizontal.



Recall that the first derivative is equal to the slope of the tangent line at a certain point. So,
If the tangent line is horizontal, then the slope is equal to 0.

We are required to solve for the points where $y' = 0$. So,

$\begin{equation} \begin{aligned} y' = \frac{d}{dx} \left[ -x^2 + 8x - 11 \right] &= 0 \\ \\ -2x + 8 &= 0 \\ \\ 2x &= 8 \\ \\ x &= 4 \end{aligned} \end{equation}$


Then by substitution,

$\begin{equation} \begin{aligned} y &= -x^2 + 8x - 11\\ \\ &= -(4)^2 + 8(4) - 11\\ \\ &= 5 \end{aligned} \end{equation}$


Therefore, the point is $(4,5)$