College Algebra, Chapter 4, 4.1, Section 4.1, Problem 24

A quadratic function $f(x) = x^2 - 8x + 8$.

a.) Find the quadratic function in standard form.


$
\begin{equation}
\begin{aligned}

f(x) =& x^2 - 8x + 8
&&
\\
\\
f(x) =& (x^2 - 8x + 16) + 8 - (1)(16)
&& \text{Complete the square: add } \displaystyle \left( \frac{8}{2} \right)^2 = 16 \text{ inside the parentheses, subtract $(1)(16)$ outside}
\\
\\
f(x) =& (x - 4)^2 - 8
&& \text{Factor and simplify}

\end{aligned}
\end{equation}
$


The standard form is $f(x) = (x - 4)^2 - 8$.

b.) Draw its graph.







c.) Find its maximum or minimum value.

Based from the graph in part (b), since the graph opens upward the minimum value of $f$ is $f(4) = -8$.