Question

this question has two parts. first, answer part a. then, answer part b.
part a
persevere write the equation ( y = |x - 3| + 2 ) as a piecewise - defined function.
( y=\begin{cases}square x+square & \text{if } xltsquare\square x+square & \text{if } xgeqsquareend{cases} ) (with some filled - in values in the original image: first row has 3, 5, 3; second row has 2, 1, 3)

Explanation:

Step1: Recall absolute value definition

The absolute value function \(|u|=

$$\begin{cases}u, & u\geq0\\ -u, & u<0\end{cases}$$

\). For \(y = |x - 3|+2\), let \(u=x - 3\).

Step2: Find intervals for \(x\)

Case 1: When \(x - 3\geq0\) (i.e., \(x\geq3\)), \(|x - 3|=x - 3\). So \(y=(x - 3)+2=x - 1\).
Case 2: When \(x - 3<0\) (i.e., \(x<3\)), \(|x - 3|=-(x - 3)=-x + 3\). So \(y=(-x + 3)+2=-x + 5\).

Step3: Write piecewise function

Thus, \(y=

$$\begin{cases}-x + 5, & x<3\\x - 1, & x\geq3\end{cases}$$

\)

Answer:

\(y=

$$\begin{cases}-x + 5, & x<3\\x - 1, & x\geq3\end{cases}$$

\)