Question

graph the piece - wise function given below.
f(x)=\begin{cases}|x|-4&\text{for }-7leq x < 4\\-x&\text{for }4leq xleq8end{cases}
step 1: select a function then move with blue dot (if necessary)

Explanation:

Step1: Analyze \(y = |x|-4\) for \(-7\leq x < 4\)

When \(x\geq0\), \(y=x - 4\). When \(x=-7\), \(y=|-7|-4=3\); when \(x = 0\), \(y=-4\); when \(x=4\), \(y=|4|-4 = 0\) (but this point is not included for this part of the piece - wise function). The graph of \(y = |x|-4\) is a V - shaped graph shifted down 4 units. For \(x\geq0\) it is a line with slope 1 and \(y\) - intercept \(-4\), and for \(x<0\) it is a line with slope - 1 and \(y\) - intercept \(-4\).

Step2: Analyze \(y=-x\) for \(4\leq x\leq8\)

When \(x = 4\), \(y=-4\); when \(x = 8\), \(y=-8\). This is a straight - line with slope \(-1\) and passing through the points \((4,-4)\) and \((8,-8)\).

Answer:

Graph \(y = |x|-4\) for \(-7\leq x < 4\) (a V - shaped graph with vertex at \((0, - 4)\) and endpoints \((-7,3)\) and an open - circle at \((4,0)\)) and \(y=-x\) for \(4\leq x\leq8\) (a straight - line segment with endpoints \((4,-4)\) and \((8,-8)\)) on the coordinate plane.