Chapter 3, Problem 29RE

To determine

Given quadratic function has a maximum value or minimum value.

Answer to Problem 29RE

Maximum value is $16$ .

Explanation of Solution

Given information:

  $f\left(x\right)=-3x^2+12x+4$

Calculation:

Compare the $-3x^2+12x+4$ with $f\left(x\right)=a{x}^2+bx+c$

So, $a=-3,b=12,c=4$

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 $a=-3,b=12$ and then evaluate $\frac{-b}{2a}$ .

So maximum value is,

  $\begin{array}{l}x=\frac{-b}{2a}\\ =\frac{-12}{2\left(-3\right)}\\ =2\end{array}$

Now evaluating function $\frac{-b}{2a}$ by putting $x=1$ in the function $f\left(x\right)=-3x^2+12x+4$ to get maximum value.

  $\begin{array}{l}f\left(\frac{-b}{2a}\right)=f\left(2\right)\\ =3{\left(2\right)}^2+12\left(2\right)+4\\ =-12+24+4\\ =16\end{array}$

Thus, maximum value is $16$ .