Q 58. - Precalculus Enhanced with Graphing Utilities

Question

Determine without graphing whether the given quadratic function has a maximum value or a minimum value and then find the value.

$f\left(x\right)=-2x^2+8x+3$

Step-by-Step Solution

Answer

The maximum value is $11$.

Step 1. Given information.

The given function is $f\left(x\right)=-2x^2+8x+3$.

Step 2. Determine whether the graph of the function opens up or down.

Compare the given function $f\left(x\right)=-2x^2+8x+3$ with $f\left(x\right)=a{x}^2+bx+c$.

So, here $a=-2,b=8,c=3$.

If $a=-2<0$, then the parabola opens down, which means that the function has a maximum value and it is the vertex.

Step 3. Determine the maximum value.

The maximum value occurs at $x=-\frac{b}{2a}$.

$\begin{array}{rcl}x& =& -\frac{8}{2\left(-2\right)}\\ & =& \frac{8}{4}\\ & =& 2\end{array}$

The required maximum value is:

$\begin{array}{rcl}f\left(x\right)& =& -2x^2+8x+3\\ & =& -2{\left(2\right)}^2+8\left(2\right)+3\\ & =& -8+16+3\\ & =& 11\end{array}$