Question

Couldn’t fit the whole problem in one picture so the rest will be listed in the comments

Answer 1

PART A

To find the solution to the given values in part A we trace each value on the x-axis to their corresponding point on the graph.

For example,

i) h(-2), we trace the value of -2 to the point where it crosses the graph. When we trace -2 this gives 0

Therefore

`h(-2)=0`

ii)h(0), we trace the value of 0 to the point where it crosses the graph. When we trace 0 this gives -2

Therefore

`h(0)=-2`

iii) h(2), we trace the value of 2 to the point where it crosses the graph. When we trace 2 this gives 2

`h(2)=2`

iv) h(3, we trace the value of 3 to the point where it crosses the graph. When we trace 3 this gives 3

`h(3)=3`

PART B

i) The domain of h is the distance covered by the graph along the horizontal axis. This gives,

`\text{Domain}=\lbrack-3,4\rbrack`

ii) The range of h is the distance covered by the graph along the vertical axis. This gives,

`\text{Range}=\lbrack-2,3\rbrack`

PART C

To find the values of h(x)= 2, we trace 2 on the vertical axis to the points it touches the graph then trace the points down to the horizontal axis.

Therefore, h(x)=2 gives the following values for x

`x=-3,2,4`

PART D

To find the values of h(x)<=2, from the graph we trace 2 on the vertical axis to the points it touches the graph then trace the points down to the horizontal axis.

But this time, since 2 would give -3 and 2 and they fall in a range the answer would be different from part C and would be written as

`\lbrack-3,2\rbrack\text{ and 4}`

PART E

To find the net change in h between x= -3 and 3 we trace the value of -3 and 3 from the horizontal axis to the graph, then to their corresponding points on the vertical axis. Then we subtract the solution to get the net change.

When x=-3 this gives 2 on the vertical axis

When x = 3 this gives 3 on the vertical axis

Therefore, the net change is 3 -2 = 1