Q 54. - 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.

$f\left(x\right)=-2x^2+12x$

Step-by-Step Solution

Answer

The maximum value is $18$.

Step 1. Given information.

The given quadratic function is $f\left(x\right)=-2x^2+12x$.

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

Compare the given quadratic function $f\left(x\right)=-2x^2+12x$ with $f\left(x\right)=a{x}^2+bx+c$.

So, $a=-2,b=12,c=0$.

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

Step 3. Determine the minimum value.

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

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

The maximum value is:

$\begin{array}{rcl}f\left(x\right)& =& -2x^2+12x\\ f\left(3\right)& =& -2{\left(3\right)}^2+12\left(3\right)\\ & =& -18+36\\ & =& 18\end{array}$

Step 4. Simplified answer.

Hence, the required maximum value is $18$.