College Algebra, Chapter 4, 4.1, Section 4.1, Problem 18

A quadratic function $f(x) = -3x^2 + 6x - 2$.

a.) Find the quadratic function in standard form.


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


The standard form is $f(x) = -3 (x - 1)^2 + 1$.

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 $f(x) = -3 (x - 1)^2 + 1$ is at $(1, 1)$.


$\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 = -3 (x - 1)^2 + 1 && \text{Add } 3 (x - 1)^2 &&& y = -3(0 - 1)^2 + 1 &&&& \text{Substitute } x = 0 \\ \\ & 3(x - 1)^2 = 1 && \text{Divide } 3 &&& y = -3 + 1 &&&& \text{Simplify} \\ \\ & (x - 1)^2 = \frac{1}{3} && \text{Take the square root} &&& y = -2 &&&& \\ \\ & x - 1 = \pm \sqrt{\frac{1}{3}} && \text{Add } 1 &&& &&&& \\ \\ & x = \pm \sqrt{\frac{1}{3}} + 1 && &&& &&&& \end{aligned} \end{equation}$


c.) Draw its graph.