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.