Question

r(x) = \

$$\begin{cases} x^2 - 4 & \\text{for } x \\leq 2 \\\\ 3x - 6 & \\text{for } x > 2 \\end{cases}$$

part 1 of 5
the first rule defines a \boxed{parabola} with vertex \boxed{(0, -4)}.
part 2 of 5
since x \leq 2, the \boxed{rightmost} point \boxed{(2, 0)} is graphed as a solid dot.
part 3 of 5
the second \boxed{line} with slope \boxed{1} and y-intercept \boxed{(0, -2)}.
try again

Explanation:

Step1: Identify line form

The second rule is $r(x)=3x-6$ for $x>2$, which follows the slope-intercept form $y=mx+b$, where $m$ is slope and $b$ is y-intercept.

Step2: Extract slope

From $y=3x-6$, the slope $m=3$.

Step3: Extract y-intercept

From $y=3x-6$, the y-intercept is $(0, -6)$.

Step4: Find leftmost point

For $x>2$, calculate $r(2)=3(2)-6=0$, so the leftmost point is an open dot at $(2,0)$.

Answer:

The second rule defines a line with slope $\boldsymbol{3}$ and y-intercept $\boldsymbol{(0, -6)}$. The leftmost point $(2, 0)$ is graphed as an open dot.