Question

graph the piece - wise defined function.
f(x)=\begin{cases}2x + 5&xleq0\\frac{1}{3}x + 1&x>0end{cases}
choose the correct graph below.

Explanation:

Step1: Analyze \(y = 2x+5\) for \(x\leq0\)

When \(x = 0\), \(y=2\times0 + 5=5\). The slope is \(m = 2\). For \(x\leq0\), we start at the point \((0,5)\) and use the slope to find other points. Since \(x\leq0\), we can take \(x=- 1\), then \(y=2\times(-1)+5 = 3\).

Step2: Analyze \(y=\frac{1}{3}x + 1\) for \(x>0\)

When \(x = 0\), the function is not defined at this value for \(y=\frac{1}{3}x + 1\) in its given domain. But as \(x\) approaches \(0\) from the right, \(y=\frac{1}{3}\times0+1 = 1\). The slope of this line is \(m=\frac{1}{3}\). We can take \(x = 3\), then \(y=\frac{1}{3}\times3+1=2\).

Step3: Check the graph features

The graph of \(y = 2x + 5\) for \(x\leq0\) should have a closed - circle at \((0,5)\) (because \(x = 0\) is included in the domain \(x\leq0\)), and the graph of \(y=\frac{1}{3}x + 1\) for \(x>0\) should have an open - circle at \((0,1)\) (because \(x = 0\) is not included in the domain \(x>0\)).

Answer:

(Without seeing the actual graphs A, B, C, D, you would look for a graph with a line \(y = 2x+5\) with a closed - circle at \((0,5)\) for \(x\leq0\) and a line \(y=\frac{1}{3}x + 1\) with an open - circle at \((0,1)\) for \(x>0\))