Question

Determine whether each function has a maximum or minimum value. Then find the maximum or minimum value of the function.

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

Short answer

The function has minimum value that is -14.

Step-by-step solution

Step 1. Use the concept.

Consider the function $f\left(x\right)=a{x}^2+bx+c,a\ne 0$, the x-coordinate of vertex is $\frac{-b}{2a}$.

The graph of $f\left(x\right)=a{x}^2+bx+c,a\ne 0$

• opens up and has a minimum value when $a>0$, and
• opens down and has a maximum value when $a<0$

Step 2. Given Information.

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

Step 3. Solution.

In the function$f\left(x\right)=x^2-8x+2$, we have $a=1,b=-8,c=2$

Here, $a=1>0$

So, the graph opens up and has a minimum value.

The minimum value of the function is the y-coordinate of the vertex.

Thex-coordinate of the vertex is

$\begin{array}{l}\frac{-b}{2a}\\ =\frac{-\left(-8\right)}{2\left(1\right)}\mathrm{..........}\left[a=1,b=-8\right]\\ =4\end{array}$

Find they-coordinate of the vertex by evaluating the function for $x=4$.

$f\left(4\right)={\left(4\right)}^2-8\left(4\right)+2=-14$

So, the minimum value of function is -14.