Chapter 3, Problem 27RE

To determine

Given quadratic function has a maximum value or minimum value.

Answer to Problem 27RE

Maximum value is $12$ .

Explanation of Solution

Given information:

  $f\left(x\right)=-x^2+8x-4$

Calculation:

Compare the $-x^2+8x-4$ with $f\left(x\right)=a{x}^2+bx+c$

So, $a=1,b=8,c=-4$

The graph of a quadratic function $f\left(x\right)=a{x}^2+bx+c$ opens up if $a>0$ and opens down if $a<0$ .

Function has $a=-1$ which is greater than $0$ .

So graph is opens down.

In case $a<0$ vertex is a highest point on the graph. Also, in case $a>0$ , then the vertex is a lowest point on the graph.

Since $a<0$ , the vertex is a maximum point.

Minimum value is,

Minimum value of the function $f$ is $f\left(\frac{-b}{2a}\right)$ if the vertex is the lowest point.

Put $a=-1,b=8$ and then evaluate $\frac{-b}{2a}$ .

So maximum value is at,

  $\begin{array}{l}x=\frac{-b}{2a}\\ =\frac{-8}{2\left(-1\right)}\\ =4\end{array}$

Now evaluating function $\frac{-b}{2a}$ by putting $x=1$ in the function $f\left(x\right)=-x^2+8x-4$ to get maximum value.

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

Thus, maximum value is $12$ .