Q 57. - 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.

$f\left(x\right)=-x^2+10x-4$

Step-by-Step Solution

Answer

The maximum value is $21$.

Step 1. Given information.

The given equation is $f\left(x\right)=-x^2+10x-4$.

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

Compare the given quadratic function $f\left(x\right)=-x^2+10x-4$ with $f\left(x\right)=a{x}^2+bx+c$.

So, here $a=-1,b=10,c=-4$.

If $a=-1<0$, 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 $x=-\frac{b}{2a}$.

$\begin{array}{rcl}x& =& -\frac{10}{2\left(-1\right)}\\ & =& \frac{10}{2}\\ & =& 5\end{array}$

The maximum value is:

$\begin{array}{rcl}f\left(x\right)& =& -x^2+10x-4\\ & =& -{\left(5\right)}^2+10\left(5\right)-4\\ & =& -25+50-4\\ & =& 25-4\\ & =& 21\end{array}$

Step 4. Simplified answer.

Hence, the required maximum value is $21$.