Question

r(x) = \

$$\begin{cases} x^2 - 4 & \\text{for } x \\leq 2 \\\\ 5x - 10 & \\text{for } x > 2 \\end{cases}$$

part 1 of 5
the first rule defines a \underline{parabola} with vertex \underline{(0, -4)}.
part 2 of 5
since x \leq 2, the \underline{rightmost} point \underline{(2, 0)} is graphed as a solid dot.
part 2 / 5
part 3 of 5
the second \underline{try one last time} \underline{\text{dropdown}} with slope \underline{5} and y-intercept \underline{(0, -4)} \text{with a red cross}
\text{buttons: 6, ×}

Explanation:

Step1: Identify line equation form

The second function is $r(x)=5x-10$ for $x>2$, which follows the slope-intercept form $y=mx+b$, where $m$ is slope and $b$ is y-intercept.

Step2: Calculate y-intercept

To find the y-intercept, set $x=0$:
$r(0)=5(0)-10=-10$
So the y-intercept is $(0, -10)$.

Step3: Identify leftmost point

For $x>2$, the leftmost point is at $x=2$ (open dot, since $x$ cannot equal 2):
$r(2)=5(2)-10=10-10=0$
So the leftmost point is $(2, 0)$.

Answer:

The second rule defines a line with slope 5 and y-intercept $(0, -10)$. Since $x>2$, the leftmost point $(2, 0)$ is graphed as an open dot.