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graph this function. $f(x)=\\begin{cases}3&\\text{if }x < - 6\\\\-\\fra…

Question

graph this function.
$f(x)=\

$$\begin{cases}3&\\text{if }x < - 6\\\\-\\frac{1}{2}x + 4&\\text{if }x\\geq - 6\\end{cases}$$

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

Explanation:

Step1: Analyze the first - part of the piece - wise function

For $f(x)=3$ when $x < - 6$, it is a horizontal line $y = 3$ for all $x$ values less than $-6$. The point at $x=-6$ for this part is open since the inequality is $x < - 6$.

Step2: Analyze the second - part of the piece - wise function

For $f(x)=-\frac{1}{2}x + 4$ when $x\geq - 6$, it is a linear function. The slope $m=-\frac{1}{2}$ and the $y$ - intercept $b = 4$. When $x=-6$, $y=-\frac{1}{2}\times(-6)+4=3 + 4=7$. The point at $x = - 6$ for this part is closed since the inequality is $x\geq - 6$.

Step3: Find other points for the linear part

Let $x = 0$, then $y=-\frac{1}{2}\times0 + 4=4$. So we have the point $(0,4)$.

To graph:

  1. Draw a horizontal line $y = 3$ for $x < - 6$ with an open - circle at $x=-6$.
  2. Draw the line $y=-\frac{1}{2}x + 4$ starting from the point $(-6,7)$ (closed - circle) and passing through $(0,4)$.

Answer:

Graph a horizontal line $y = 3$ for $x < - 6$ with an open - circle at $(-6,3)$ and a line $y=-\frac{1}{2}x + 4$ starting from the point $(-6,7)$ (closed - circle) and passing through $(0,4)$.