Please help me with the domain and range of the graphs.?part 5

Problem 5

Domain =

This is because the smallest x value allowed is x = -4 as shown by the left-most point. The right most point has x coordinate x = 3, so this is the largest value in the domain.

The domain in interval notation is [-4, 3]. The square brackets include the endpoint.

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Range =

We're now looking at the possible y values. The lowest point happens when y = -4 and the highest point is when y = 5. We can have any y value between those endpoints.

The range in interval notation is [-4, 5]

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Problem 6

Domain =

Same idea as the previous problem. This time the left most point has x = -2 and the right most point has x coordinate x = 2.

The big difference here is that the open hole says to not include the endpoint. The domain in interval notation is [-2, 2) where we use a parenthesis to not include the endpoint.

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Range =

The smallest possible y value is y = -4. The ceiling for the y values is y = 4, but we can't actually reach this value due to the open hole.

The range in interval notation is [-4, 4)

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Problem 7

Domain =

Same idea as earlier. Now the open hole is at the left endpoint. The right most point occurs not at an endpoint this time.

The domain in interval notation is (-4, 2]

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Range:

Both endpoints are open holes, so we exclude both endpoints.

The range in interval notation is (-4, 4). Unfortunately in this case, interval notation looks identical to ordered pair notation.

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Problem 8

Domain =

Domain in interval notation = (-5, 3)

The graph is too blurry to be able to determine the range.

table 3 represents a direct variation function. : )

correct choice is b

step-by-step explanation:

when you calculate the average rate of change of a function, you are finding the slope of the secant line between the two points. for the function the average rate of change from point to point can be calculated as

to find the average rate of change from day 2 to day 10, note that at day 2, 8 houses were sold, so a(2,8) and at day 10, 2 houses were sold, so b(10,2). thus,