iv. graph the following functions and label the critical points: 7. f(x)=-3|x - 2|+6 8. f(x)=2|x + 4|-2

Explanation:

Step1: Recall definition of absolute - value function

The absolute - value function (y = |u|) can be written as (y=\begin{cases}u, & u\geq0\-u, & u < 0\end{cases}). For (y=-3|x - 2|+6), when (x-2\geq0) (i.e., (x\geq2)), (y=-3(x - 2)+6=-3x+6 + 6=-3x + 12); when (x-2<0) (i.e., (x<2)), (y=-3(-(x - 2))+6 = 3x-6 + 6=3x).

Step2: Find critical points

The derivative of (y = 3x) for (x<2) is (y'=3), and the derivative of (y=-3x + 12) for (x>2) is (y'=-3). The function (y=-3|x - 2|+6) is not differentiable at (x = 2). To find the (y) - value at (x = 2), substitute (x = 2) into (y=-3|x - 2|+6), we get (y = 6). So the critical point is ((2,6)).

Step3: Analyze the second function

For (y = 2|x + 4|-2), when (x+4\geq0) (i.e., (x\geq - 4)), (y=2(x + 4)-2=2x+8 - 2=2x+6); when (x+4<0) (i.e., (x<-4)), (y=2(-(x + 4))-2=-2x-8 - 2=-2x-10).

Step4: Find critical points of the second function

The derivative of (y=-2x - 10) for (x<-4) is (y'=-2), and the derivative of (y = 2x+6) for (x>-4) is (y'=2). The function (y = 2|x + 4|-2) is not differentiable at (x=-4). Substitute (x=-4) into (y = 2|x + 4|-2), we get (y=-2). So the critical point is ((-4,-2)).

Step5: Graph the functions

For (y=-3|x - 2|+6), the vertex is at ((2,6)), the slope to the left of (x = 2) is (3) and to the right of (x = 2) is (-3). For (y = 2|x + 4|-2), the vertex is at ((-4,-2)), the slope to the left of (x=-4) is (-2) and to the right of (x=-4) is (2).

Answer:

For (f(x)=-3|x - 2|+6), the critical point is ((2,6)). For (f(x)=2|x + 4|-2), the critical point is ((-4,-2)).