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.