Q. 17 - 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.

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

Step-by-Step Solution

Answer

The quadratic function has a maximum value. The maximum value is 4.

Step 1. Given information.

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

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

Step 2. Compare the function.

Compare

$$\begin{gathered} f\left(x\right)=-2x^2+4 \mathrm{to} f\left(x\right)=a{x}^2+bx+c \\ \mathrm{Then} \mathrm{a}=-2, \mathrm{b}=0 \mathrm{and} \mathrm{c}=4 \end{gathered}$$

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

$x=-\frac{b}{2a}=-\frac{0}{2\left(-2\right)}=0$

The maximum value is 

$f\left(-\frac{b}{2a}\right)=f\left(0\right)=-2{\left(0\right)}^2+4=4$

Step 3. Conclusion.

The quadratic function has a maximum value. The maximum value is 4.