College Algebra, Chapter 4, 4.1, Section 4.1, Problem 20

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

a.) Find the quadratic function in standard form.


$
\begin{equation}
\begin{aligned}

f(x) =& 2x^2 + x - 6
&&
\\
\\
f(x) =& 2 \left( x^2 + \frac{1}{2} x \right) - 6
&& \text{Factor out $2$ from $x$-terms}
\\
\\
f(x) =& 2 \left( x^2 + \frac{1}{2} x + \frac{1}{16} \right) - 6 - (2) \left( \frac{1}{16} \right)
&& \text{Complete the square: add } \frac{1}{16} \text{ inside parentheses, subtract } (2)\left( \frac{1}{16} \right) \text{ outside}
\\
\\
f(x) =& 2 \left( x + \frac{1}{4} \right)^2 - \frac{49}{8}
&& \text{Factor and simplify}

\end{aligned}
\end{equation}
$


The standard form is $\displaystyle f(x) = 2 \left( x + \frac{1}{4} \right)^2 - \frac{49}{8}$.

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 $\displaystyle f(x) = 2 \left( x + \frac{1}{4} \right)^2 - \frac{49}{8} $ is at $\displaystyle \left( \frac{-1}{4}, \frac{-49}{8} \right)$.


$
\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 = 2 \left(x + \frac{1}{4} \right)^2 - \frac{49}{8}
&& \text{Add } \frac{49}{8}
&&& y = 2 \left( 0 + \frac{1}{4} \right)^2 - \frac{49}{8}
&&&& \text{Substitute } x = 0
\\
\\
& \frac{49}{8} = 2 \left( x + \frac{1}{4} \right)^2
&& \text{Divide } 2
&&& y = \frac{1}{8} - \frac{49}{8}
&&&& \text{Simplify}
\\
\\
& \frac{49}{16} = \left(x + \frac{1}{4} \right)^2
&& \text{Take the square root}
&&& y = -6
&&&&
\\
\\
& \pm \frac{7}{4} = x + \frac{1}{4}
&& \text{Subtract } \frac{1}{4}
&&&
&&&&
\\
\\
& x = \pm \frac{7}{4} - \frac{1}{4}
&&
&&&
&&&&
\\
\\
& x = -2 \text{ and } x = \frac{3}{2}
&&
&&&
&&&&


\end{aligned}
\end{equation}
$


c.) Draw its graph.