College Algebra, Chapter 4, 4.1, Section 4.1, Problem 10

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

a.) Find the quadratic function in standard form.


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


The standard form is $f(x) = (x + 4)^2 - 16$.

b.) Find its vertex and its $x$ and $y$-intercepts.

Using the formula of standard form of a Quadratic function,

$f(x) = a (x - h)^2 + k$

We know that the vertex is at $(h, k)$, so the vertex of $f(x) = (x + 4)^2 - 16$ is at $(-4, -16)$.


$\begin{array}{llll}
\text{Solving for $x$-intercepts} & & \text{Solving for $y$-intercept} & \\
\text{We set } f(x) = 0, \text{then} & & \text{We set } x = 0, \text{ then} & \\
0 = (x + 4)^2 - 16 & \text{Add } 16 & y = (0 + 4)^2 - 16 & \text{Substitute } x = 0 \\
16 = (x + 4)^2 & \text{Take the square root} & y = 16 - 16 & \text{Simplify} \\
\pm 4 = x + 4 & \text{Subtract } 4 & y = 0 & \\
x = \pm 4 - 4 & \text{Simplify} & & \\
x = 0 \text{ and } x = -8 & & &
\end{array}
$


c.) Draw its graph.