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.