Question

In Problems 15–17, determine whether the given quadratic function has a maximum value or a minimum value, and then find the value.

$f\left(x\right)=3x^2-6x+4$

Short answer

The quadratic function has a minimum value. The minimum value is$1$.

Step-by-step solution

Step 1. Given information.

Determine whether the given quadratic function has a maximum value or a minimum value, and then find the value.

$f\left(x\right)=3x^2-6x+4$

Step 2. Compare the function.

Compare $$\begin{gathered} f\left(x\right)=3x^2-6x+4\mathrm{to}f\left(x\right)=a{x}^2+bx+c \\ \mathrm{Then}\mathrm{a}=3,\mathrm{b}=-6\mathrm{and}\mathrm{c}=4 \end{gathered}$$

Since $a>0$, the graph of f opens up, so the vertex is a minimum value occurs at

$x=-\frac{b}{2a}=-\frac{-6}{2\left(3\right)}=\frac{6}{6}=1$

The minimum value is

$f\left(-\frac{b}{2a}\right)=f\left(1\right)=3{\left(1\right)}^2-6\left(1\right)+4=3-6+4=1$

Step 3. Conclusion.

The quadratic function has a minimum value. The minimum value is$1$.