Question

consider the piecewise h(x) = -2/3(x+1) x<-1
-4x-4 -1≤ x≤ 0
x-8 1 11 years ago

Answer 1

@aum

Answer 2

For x < -1, h(x) = -2/3(x + 1). This is a straight line. We need two points on the line. Pick x = -4 and x = -7 and find the y-values:
h(x) = -2/3(x + 1)
x = -4, h(x) = -2/3(-4+1) = -2/3 * (-3) = 2. (-4, 2) is a point
x = -7, h(x) = -2/3(-7+1) = -2/3 * (-6) = 4. (-7, 4) is another point.
Plot the two points and draw a line through them but STOP at x = -1 because this part of the equation is valid only for x < -1.

Answer 3

The second part is: h(x) = -4x - 4 for -1 <= x <= 0
This is also a straight line. Pick two points in the interval [-1, 0]
Pick x = -1 and x = 0
x = -1, h(x) = -4(-1) - 4 = 4 - 4 = 0. (-1, 0) is one point.
x = 0, h(x) = -4(0) - 4 = 0 - 4 = -4. (0, -4) is one point.
Plot the two points and draw a line through them but STOP at x = -1 at the left end and x = 0 at the right end because this part of the equation is valid only for -1 <= x <= 0.

Answer 4

Can you do the third part yourself using the method as shown above?

Answer 5

Click on the link and then click on the graph for a bigger version:
http://awesomescreenshot.com/0303lr6g9d

Answer 6

When you see a open circle, that point is NOT included in that line.
When you see a closed circle, that point IS included in that line.

Answer 7

so how do you find the range?

Answer 8

If you look at the graph, the lowest value of the function is at y = -7 (but -7 is not included as indicated by the open circle at the bottom of the green line). The highest value is the left side of the red line which will keep going towards positive infinity. Therefore, the range is (-7, infinity).

Answer 9

but doesn't the left side go to negative infinity

Answer 10

As the x-value goes to negative infinity, the y-value goes to positive infinity. Here they are asking for the range and so we are interested in the y-values. We can say the range is (-7, infinity) in interval notation or we can simply say y > -7 is the range.