Question

graph the function.
( r(x) = \begin{cases} x^2 - 4 & \text{for } x leq 2 \\ 3x - 6 & \text{for } x > 2 end{cases} )
part: 0 / 5
part 1 of 5
the first rule defines a
select
with vertex

Explanation:

Step1: Identify first rule's function type

The first rule is $r(x)=x^2-4$ for $x\leq2$, which is a quadratic function in the form $y=x^2+k$.

Step2: Find vertex of quadratic function

For $y=x^2-4$, the vertex form is $y=(x-0)^2-4$. The vertex of $y=(x-h)^2+k$ is $(h,k)$. Here $h=0$, $k=-4$.

Answer:

The vertex is $(0, -4)$

To complete the graphing process (for full context):

1. For $r(x)=x^2-4$ ($x\leq2$): Plot the vertex $(0,-4)$, find additional points e.g., $x=2$ gives $r(2)=2^2-4=0$, so $(2,0)$ (closed dot, since $x\leq2$), $x=-2$ gives $r(-2)=(-2)^2-4=0$, so $(-2,0)$, then draw the left half and up to $x=2$ of the parabola.
2. For $r(x)=3x-6$ ($x>2$): Find the point at $x=2$: $r(2)=3(2)-6=0$ (open dot, since $x>2$), find another point e.g., $x=3$ gives $r(3)=3(3)-6=3$, so $(3,3)$, then draw the line starting from the open dot at $(2,0)$ and extending right.