Q. 16 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 15–17, determine whether the given quadratic function has a maximum value or a minimum value, and then find the value.

$f\left(x\right) =-x^2+8x-4$

Step-by-Step Solution

Answer

The quadratic function has a maximum value. The maximum value is 12.

Step 1. Given information.

Determine whether the given quadratic function has a maximum value or a minimum value, and then find the value. 

$f\left(x\right)=-x^2+8x-4$

Step 2. Compare the function.

Compare

$$\begin{gathered} f\left(x\right)=-x^2+8x-4 \mathrm{to} f\left(x\right)=a{x}^2+bx+c \\ \mathrm{Then} \mathrm{a}=-1, \mathrm{b}=8 \mathrm{and} \mathrm{c}=-4 \end{gathered}$$

Since$a<0$ the graph of f opens down, so the vertex is a maximum value occurs at

$x=-\frac{b}{2a}=-\frac{8}{2(-1)}=4$

The maximum value is 

$f(-\frac{b}{2a})=f\left(4\right)=-4^2+8\left(4\right)-4=-16+32-4=12$

Step 3. Conclusion.

The quadratic function has a maximum value. The maximum value is 12.