The graph of y = h(x) is a line segment joining the points (1, -5) and (9,1). Drag the endpoints of the segment below to graph y = h-'(x).

Ok, i cant drag the endpoints of the segment, but i can tell you how to do it.

First, we know that h(x) joins the points (1, -5) and (9, 1), then h(x) is a line:

h(x) = s*x + b

First, for a line that goes through the points (x1, y1) and (x2, y2), the slope will be:

s = (y2 -y1)/(x2 - x1)

Then in this case, the slope is:

s = (1 - (-5))/(9 - 1) = 0.75

Then we have

h(x) = 0.75*x + b

now, the value of b can be found as:

h(1) = -5 = 0.75*1 - b

b = - 5 - 0.75 = -5.75.

Then our equation is:

h(x) = 0.75*x - 5.75

Now, i gues you want to find the graph of:

y = h(-x)

Then our new function is:

g(x) = h(-x) =  -0.75*x - 5.75.

Now to find the points, we evaluate this function in the same values of x as before.

g(1) = -0.75*1 - 5,75 = -6,5

the point is (1, -6.5)

the second point is when x = 9.

g(9) = -0.75*9 - 5.75 = -12.5

The second point is (9, -12.5)