SOLUTION
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
[tex]h(-2)=0[/tex]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
[tex]h(0)=-2[/tex]iii) h(2), we trace the value of 2 to the point where it crosses the graph. When we trace 2 this gives 2
[tex]h(2)=2[/tex]iv) h(3, we trace the value of 3 to the point where it crosses the graph. When we trace 3 this gives 3
[tex]h(3)=3[/tex]PART B
i) The domain of h is the distance covered by the graph along the horizontal axis. This gives,
[tex]\text{Domain}=\lbrack-3,4\rbrack[/tex]ii) The range of h is the distance covered by the graph along the vertical axis. This gives,
[tex]\text{Range}=\lbrack-2,3\rbrack[/tex]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
[tex]x=-3,2,4[/tex]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
[tex]\lbrack-3,2\rbrack\text{ and 4}[/tex]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