College Algebra, Chapter 4, 4.1, Section 4.1, Problem 22

A quadratic function $f(x) = 6x^2 + 12x - 5$.

a.) Find the quadratic function in standard form.


$
\begin{equation}
\begin{aligned}

f(x) =& 6x^2 + 12x - 5
&&
\\
\\
f(x) =& 6 (x^2 + 2x) - 5
&& \text{Factor out $6$ from $x$-terms}
\\
\\
f(x) =& 6 (x^2 + 2x + 1) - 5 - (6)(1)
&& \text{Complete the square: add $1$ inside parentheses, subtract $(6)(1)$ outside}
\\
\\
f(x) =& 6 (x + 1)^2 - 11
&& \text{Factor and simplify}

\end{aligned}
\end{equation}
$


The standard form is $\displaystyle f(x) = 6 (x + 1)^2 - 11$.

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 $f(x) = 6 (x + 1)^2 - 11$ is at $(-1, -11)$.


$
\begin{equation}
\begin{aligned}

& \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 = 6 (x + 1)^2 - 11
&& \text{Add } 11
&&& y = 6 (0 + 1)^2 - 11
\\
\\
& 11 = 6 (x + 1)^2
&& \text{Divide } 6
&&& y = 6 - 11
\\
\\
& \frac{11}{6} = (x + 1)^2
&& \text{Take the square root}
&&& y = -5
\\
\\
& \pm \sqrt{\frac{11}{6}} = x + 1
&& \text{Subtract } 1
&&&
\\
\\
& x = \pm \sqrt{\frac{11}{6}} - 1
&&
&&&

\end{aligned}
\end{equation}
$


c.) Draw its graph.