Question

Determine without graphing whether the given quadratic function has a maximum value or a minimum value and then find the value.

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

Short answer

The minimum value is$-1$.

Step-by-step solution

Step 1. Given information.

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

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

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

So, here $a=4,b=-4,c=0$.

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 $x=-\frac{b}{2a}$.

$\begin{array}{rcl}x& =& -\frac{-4}{2\left(4\right)}\\ & =& \frac{1}{2}\end{array}$

The required minimum value is:

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

Step 4. Simplified answer.

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