Question

graph the following function on the axes provided. $f(x)=\begin{cases}-2x + 4& \text{for }x < - 1\\-5x + 19& \text{for }x>3end{cases}$ click and drag to make a line. click the line to delete it. click on an endpoint of a line to change it.

Explanation:

Step1: Find points for $y = - 2x+4$ when $x < - 1$

Let's find the value of the function at $x=-1$. Substitute $x = - 1$ into $y=-2x + 4$, we get $y=-2\times(-1)+4=6$. Since $x < - 1$, we can choose $x=-2$, then $y=-2\times(-2)+4=8$. The line $y = - 2x+4$ has a slope of $-2$ and a $y$-intercept of $4$, but we only consider the part of the line for $x < - 1$, and we draw an open - circle at the point $(-1,6)$ (because $x=-1$ is not included in the domain $x < - 1$).

Step2: Find points for $y=-5x + 19$ when $x>3$

Let's find the value of the function at $x = 3$. Substitute $x = 3$ into $y=-5x + 19$, we get $y=-5\times3+19=4$. Since $x>3$, we can choose $x = 4$, then $y=-5\times4+19=-1$. The line $y=-5x + 19$ has a slope of $-5$ and a $y$-intercept of $19$, but we only consider the part of the line for $x>3$, and we draw an open - circle at the point $(3,4)$ (because $x = 3$ is not included in the domain $x>3$).

Answer:

Graph the line $y=-2x + 4$ for $x < - 1$ (open - circle at $(-1,6)$) and the line $y=-5x + 19$ for $x>3$ (open - circle at $(3,4)$) on the given coordinate axes.