Question

PLEASE HELP!!!!

Answer 1

As per the description of my friend's work, all the steps except the last one are correct.

Step 1 is correct about the slope, the point and the filled circle. We will have the filled circle because the point x=-1 is included too.

Step 2 is correct about the two points with the open circle and that will give us a negative slope. The circle will be open as the two end points are not included.

Step 3 is partially correct in the sense that the "horizontal line to the right" part is correct. What is not correct is the "open circle on (4,2)" part. The circle at x=-4 has to be closed because, $x\geq-4$, means that the point x=-4 too is included and therefore, there has to be a filled in circle there.

Answer 2

PART 1):

f(x) = 3x – 5 if x < -1

Start with a filled in circle on (-1, -8). Use the slope to go up 3 and right 1 to make another point and drawn the line only going down.

Answer: WRONG.

Because correct inequality is x≤-1 not x<-1. Hence we will put a close circle not an open circle. at (-1,-8).

Also since slope is positive so graph will go upward not downward.

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PART 2):

f(x) = -2x + 3 if -1 < x < 4

Put an open circle on (-1,5) and open circle on (4, -5). Connect those points.

Answer: Correct

Because correct inequality is < or > so both end points will be open circles.

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PART 3):

f(x) = 2 if x > 4

Put an open in circle on (4,2) then draw a horizontal line to the right.

Answer:  WRONG.

Because correct inequality is x≥4 not x>4. Hence we will put a close circle not an open circle. at (4,2).