Chapter 3, Problem 16RE

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

16. $f(x) = -x^2 + 8x - 4$

To determine

To calculate: Whether the given function is having a maximum value or a minimum value and then finds the value.

Answer to Problem 16RE

The given has a minimum value which is at $y=12$.

Explanation of Solution

Given:

The given function is $f(x)=-x^2+8x-4$.

Formula used:

The general form of a quadratic equation is $f(x)=a{(x-h)}^2+k,a\ne 0$.

This general equation can also be written as $f(x)=a{x}^2+bx+c,a\ne 0$, where

$h=-\frac{b}{2a}$

and

$k=f\left(-\frac{b}{2a}\right)$

If $a>0$, the graph opens upwards.

If $a<0$, the graph opens downwards.

The vertex of the above function is $(h,k)=\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$.

If $a<0$, the vertex is the highest point on the graph and $f\left(-\frac{b}{2a}\right)$ is the maximum value of $f$.

If $a>0$, the vertex is the lowest point on the graph and $f\left(-\frac{b}{2a}\right)$ is the minimum value of $f$.

Calculation:

The given function is $f(x)=-x^2+8x-4$.

Here, we have $a=-1$, $b=8$ and $c=-4$.

Therefore, we have $a<0$ and thus the vertex is the highest point on the graph and the maximum value is

$f\left(-\frac{b}{2a}\right)=f\left(-\frac{8}{2(-1)}\right)=f(4)$

$=-{(4)}^2+8(4)-4$

$=-16-32-4=12$