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.