College Algebra, Chapter 4, 4.1, Section 4.1, Problem 14

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

a.) Find the quadratic function in standard form.


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


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

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


$\begin{array}{llll}
\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 = (x - 1)^2 + 1 & \text{Subtract 1} & y = (0 - 1)^2 + 1 & \text{Substitute } x = 0 \\
-1 = (x - 1)^2 & \text{Take the square root} & y = 1 + 1 & \text{Simplify} \\
\pm \sqrt{-1} = x - 1 & & y = 2 & \\
\text{$x$-intercept does not exist} & & &
\end{array}
$


c.) Draw its graph.