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$Graph the function below. Then, ic discontinuities. f(x)={[-(1)/(2)x-4," if "x < -1],[(x-2)^(2)," if "-1 <= x < 3],[|x-5|-3," if "x >= 3]:}$

$6$ . Graph the function below. Then, ic discontinuities. $\newline$ $f(x)=\left\{\begin{array}{cl} -\frac{1}{2} x-4 & \text { if } x<-1 \\ (x-2)^{2} & \text { if }-1 \leq x<3 \\ |x-5|-3 & \text { if } x \geq 3 \end{array}\right.$

Full solution

6

. Graph the function below. Then, ic discontinuities.

\newline

f(x)=\left\{\begin{array}{cl} -\frac{1}{2} x-4 & \text { if } x<-1 \\ (x-2)^{2} & \text { if }-1 \leq x<3 \\ |x-5|-3 & \text { if } x \geq 3 \end{array}\right.

Graph Piecewise Function $x < -1$ : Graph the piecewise function for $x < -1$ , which is $f(x) = -\frac{1}{2}x - 4$ . This is a straight line with a negative slope, starting from the y-intercept at $-4$ .
Graph Piecewise Function $-1 \leq x < 3$ : Graph the piecewise function for $-1 \leq x < 3$ , which is $f(x) = (x - 2)^2$ . This is a parabola opening upwards with the vertex at $(2, 0)$ .
Graph Piecewise Function $x \geq 3$ : Graph the piecewise function for $x \geq 3$ , which is $f(x) = |x - 5| - 3$ . This is a V-shaped graph with the vertex at $(5, -3)$ .
Identify Discontinuity at $x = -1$ : Identify discontinuities by checking the endpoints of the intervals. At $x = -1$ , the left-hand limit is $-\left(\frac{1}{2}\right)(-1) - 4 = -3.5$ , and the right-hand limit is $(-1 - 2)^2 = 9$ . Since these don't match, there's a discontinuity at $x = -1$ .
Identify Discontinuity at $x = 3$ : Check for discontinuity at $x = 3$ . The left-hand limit is $(3 - 2)^2 = 1$ , and the right-hand limit is $|3 - 5| - 3 = 2 - 3 = -1$ . Since these don't match, there's a discontinuity at $x = 3$ .