Question

r(x) = \

$$\begin{cases} x^2 - 1 & \\text{for } x \\leq 1 \\\\ 5x - 5 & \\text{for } x > 1 \\end{cases}$$

part 1 of 5
the first rule defines a \boxed{parabola} with vertex \boxed{(0, -1)}.
part 2 of 5
since x \leq 1, the \boxed{rightmost} point \boxed{(1, 0)} is graphed as a solid dot.
part: 2 / 5
part 3 of 5
the second \boxed{try again}
select \boxed{} with slope \boxed{} and y-intercept \boxed{(\boxed{}, \boxed{})}.

Explanation:

Step1: Identify line form

The second rule is $r(x)=5x-5$ for $x>1$, which follows the slope-intercept form $y=mx+b$, where $m$ is slope and $b$ is the y-intercept.

Step2: Extract slope

From $r(x)=5x-5$, the slope $m=5$.

Step3: Calculate y-intercept

Set $x=0$ in $r(x)=5x-5$:
$r(0)=5(0)-5=-5$, so the y-intercept is $(0,-5)$.

Answer:

Slope: $5$, y-intercept: $(0, -5)$