Question

graph the function.
f(x) = {3x + 2, x < -2; x, -2 ≤ x ≤ 3; 2x - 1, x > 3}

Explanation:

Step1: Analyze \(y = 3x+2\) for \(x < - 2\)

Find two - point form. When \(x=-3\), \(y=3\times(-3)+2=-9 + 2=-7\). The line \(y = 3x + 2\) has a slope of \(3\) and a \(y\) - intercept of \(2\), but we only consider the part where \(x < - 2\). The point \((-3,-7)\) is on this part of the line, and we draw an open - circle at \(x=-2\), \(y=3\times(-2)+2=-4\).

Step2: Analyze \(y = x\) for \(-2\leq x\leq3\)

When \(x=-2\), \(y=-2\); when \(x = 3\), \(y = 3\). We draw a solid line segment connecting the points \((-2,-2)\) and \((3,3)\) since the inequality includes the endpoints.

Step3: Analyze \(y=2x - 1\) for \(x>3\)

When \(x = 4\), \(y=2\times4-1=7\). The line \(y = 2x-1\) has a slope of \(2\) and a \(y\) - intercept of \(-1\), and we draw an open - circle at \(x = 3\), \(y=2\times3-1=5\).

To graph the function, we combine these three parts on the same coordinate plane.

Answer:

The graph consists of a part of the line \(y = 3x+2\) for \(x < - 2\) (open - circle at \(x=-2\)), a line segment \(y = x\) for \(-2\leq x\leq3\) (solid endpoints), and a part of the line \(y=2x - 1\) for \(x>3\) (open - circle at \(x = 3\)).