Question

f(x)=
\begin{cases}3x + 2&\text{if }x<5\\-3x - 2&\text{if }xgeq5end{cases}

Explanation:

Step1: Analyze the function for $x < 5$

For $f(x)=3x + 2$ when $x < 5$. The $y$-intercept is found by setting $x = 0$, so $f(0)=3(0)+2=2$. The slope is $m = 3$. This is a line with a positive slope for $x$-values less than 5.

Step2: Analyze the function for $x\geq5$

For $f(x)=-3x - 2$ when $x\geq5$. The value of the function at $x = 5$ is $f(5)=-3(5)-2=-15 - 2=-17$. The slope of this part of the function is $m=-3$, a negative - slope line for $x$-values 5 and greater.

Step3: Check the continuity and open/closed points

The function $y = 3x+2$ for $x < 5$ approaches $3(5)+2=15 + 2 = 17$ as $x$ approaches 5 from the left. The function $y=-3x - 2$ has a value of - 17 at $x = 5$. So there is a break in the graph at $x = 5$. For $x < 5$, the graph of $y = 3x+2$ has an open - circle at the point where $x = 5$ (since $x$ is strictly less than 5), and for $x\geq5$, the graph of $y=-3x - 2$ has a closed - circle at $x = 5$.

Answer:

The correct graph is the one that has a line with a positive slope $y = 3x+2$ for $x < 5$ with an open - circle at the point corresponding to $x = 5$, and a line with a negative slope $y=-3x - 2$ for $x\geq5$ with a closed - circle at the point corresponding to $x = 5$. Without seeing the exact details of each option in a,b,c,d, based on the above analysis, you can identify the correct graph.