Sunday, August 26, 2012

College Algebra, Chapter 4, 4.1, Section 4.1, Problem 28

A quadratic function $f(x) = 1 - 6x - x^2$.

a.) Find the quadratic function in standard form.


$
\begin{equation}
\begin{aligned}

f(x) =& 1 - 6x - x^2
&&
\\
\\
f(x) =& -1(x^2 + 6x) + 1
&& \text{Factor out -1 from the $x$-term}
\\
\\
f(x) =& -1 (x^2 + 6x + 9) + 1 - (-1)(9)
&& \text{Complete the square: add $9$ inside the parentheses, subtract $(-1)(9)$ outside}
\\
\\
f(x) =& - (x + 3)^2 + 10
&& \text{Factor and simplify}

\end{aligned}
\end{equation}
$


The standard form is $f(x) = - (x + 3)^2 + 10$.

b.) Draw its graph.







c.) Find its maximum or minimum value.

Based from the graph, since the graph opens downward the maximum value of $f$ is $f(-3) = 10$.

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