College Algebra, Chapter 4, 4.1, Section 4.1, Problem 16

A quadratic function $f(x) = -x^2 - 4x + 4$.

a.) Find the quadratic function in standard form.


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


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

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

By using $f(x) = a (x - h)^2 + k$ with vertex at $(h,k)$.

The vertex of the function $f(x) = - (x + 2)^2 + 8$ is at $(-2, 8)$.


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


c.) Draw its graph.