Q 59. - 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.

$f\left(x\right)=-3x^2+12x+1$

Step-by-Step Solution

Answer

The maximum value is $13$.

Step 1. Given information.

The given function is $f\left(x\right)=-3x^2+12x+1$.

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

Compare the given function $f\left(x\right)=-3x^2+12x+1$ with $f\left(x\right)=a{x}^2+bx+c$.

So, $a=-3,b=12,c=1$.

If $a=-3<0$, then 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 $x=-\frac{b}{2a}$.

$\begin{array}{rcl}x& =& -\frac{12}{2\left(-3\right)}\\ & =& \frac{12}{6}\\ & =& 2\end{array}$

The required maximum value is:

$\begin{array}{rcl}f\left(x\right)& =& -3x^2+12x+1\\ f\left(2\right)& =& -3{\left(2\right)}^2+12\left(2\right)+1\\ & =& -12+24+1\\ & =& 13\end{array}$

Step 4. Simplified answer.

Hence, the required maximum value is $13$.