Q 56. - 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.

$f\left(x\right)=4x^2-8x+3$

Step-by-Step Solution

Answer

The minimum value of the function is $-1$.

Step 1. Given information.

The given function is $f\left(x\right)=4x^2-8x+3$.

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

Compare the given function $f\left(x\right)=4x^2-8x+3$ with $f\left(x\right)=a{x}^2+bx+c$.

So, here $a=4,b=-8,c=3$.

If $a=4>0$, then the parabola opens up, which means that the function has a minimum value and it is the vertex.

Step 3. Determine the minimum value.

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

$\begin{array}{rcl}x& =& -\frac{-8}{2\times 4}\\ & =& \frac{8}{8}\\ & =& 1\end{array}$

The required minimum value is:

$\begin{array}{rcl}f\left(x\right)& =& 4x^2-8x+3\\ f\left(1\right)& =& 4{\left(1\right)}^2-8\times 1+3\\ & =& 4-8+3\\ & =& 7-8\\ & =& -1\end{array}$

Step 4. Simplified answer.

Hence, the required minimum value is $-1$.