Q 53. - 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 given quadratic equation has a minimum value that is $-18$.

Step 1. Given information.

The given quadratic equation 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 up, which means that the function has a minimum value and it is vertex.

Hence, the given quadratic function has a minimum value.

Step 3. Determine the minimum value.

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

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

The required minimum 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 minimum value is $-18$.