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 for the line. 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 \(x < - 2\). We draw an open - circle at the point when \(x=-2\), \(y=3\times(-2)+2=-6 + 2=-4\) since \(x=-2\) is not included in this part of the domain.

Step2: Analyze \(y=x\) for \(-2\leqslant x\leqslant3\)

When \(x=-2\), \(y=-2\) and when \(x = 3\), \(y = 3\). We draw a line segment connecting the points \((-2,-2)\) and \((3,3)\) with closed - circles at both endpoints since the endpoints are included in the domain.

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

When \(x = 4\), \(y=2\times4-1=7\). We draw an open - circle at the point when \(x = 3\), \(y=2\times3-1=5\) since \(x = 3\) is not included in this part of the domain and then draw the line for \(x>3\).

To graph the function, we plot the points and draw the appropriate line segments and rays according to the above - mentioned rules.

Answer:

The graph consists of a ray for \(y = 3x+2\) with an open - circle at \(x=-2\), a line segment for \(y=x\) from \(x=-2\) to \(x = 3\) with closed - circles at the endpoints, and a ray for \(y = 2x-1\) with an open - circle at \(x = 3\).