Question

graph this function.
$f(x) = \

$$\begin{cases} -1 & \\text{if } x \\leq -1 \\\\ -4x - 5 & \\text{if } x > -1 \\end{cases}$$

$
select points on the graph to plot them. select \point fill\ to change a point from closed to open.

Explanation:

Step1: Analyze first piece of function

For $f(x) = -1$ where $x \leq -1$:

• This is a horizontal line at $y=-1$.
• At $x=-1$, the point $(-1, -1)$ is closed (since $x \leq -1$ includes $x=-1$).
• We can plot another point, e.g., $(-2, -1)$ (since $-2 \leq -1$) to define the line extending left.

Step2: Analyze second piece of function

For $f(x) = -4x - 5$ where $x > -1$:

• First, find the value at the boundary $x=-1$ (open point):

$f(-1) = -4(-1) - 5 = 4 - 5 = -1$. So the point $(-1, -1)$ is open for this piece.

• Find a second point: let $x=0$, $f(0) = -4(0) -5 = -5$, so plot $(0, -5)$.
• Find a third point: let $x=1$, $f(1) = -4(1) -5 = -9$, but this is below the graph, so use $x=-0.5$: $f(-0.5) = -4(-0.5)-5=2-5=-3$, so plot $(-0.5, -3)$.

Step3: Plot and connect points

• For $x \leq -1$: Draw a horizontal line from $(-1, -1)$ (closed) leftwards, passing through points like $(-2, -1)$.
• For $x > -1$: Draw a line from $(-1, -1)$ (open) through $(0, -5)$ and $(-0.5, -3)$, extending rightwards.

Answer:

• The graph has a horizontal segment $y=-1$ for all $x \leq -1$ (closed endpoint at $(-1, -1)$), and a line $y=-4x-5$ for all $x > -1$ (open endpoint at $(-1, -1)$), passing through points such as $(0, -5)$ and $(-0.5, -3)$.