Question

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).

Answer 1

`h^(-1)(x) = (4)/(3)x + (23)/(3)`

Explanation:

Given

Graph h:

`(x_1,y_1) = (1,-5)`

`(x_2,y_2) = (9,1)`

Required

Plot
`h^(-1)(x)`

First, calculate h(x)

Calculate slope (m)

`m = (y_2 - y_1)/(x_2 - x_1)`

`m = (1--5)/(9-1)`

`m = (6)/(8)`

`m = (3)/(4)`

The equation is:

`y = m(x - x_1) + y_1`

So, we have:

`y = (3)/(4)(x - 1) -5`

`y = (3)/(4)x - (3)/(4) -5`

`y = (3)/(4)x + (-3 - 20)/(4)`

`y = (3)/(4)x - (23)/(4)`

Next, calculate
`h^(-1)(x)`

Swap y and x

`x = (3)/(4)y - (23)/(4)`

Solve for y

`(3)/(4)y = x + (23)/(4)`

Multiply through by 4

`3y = 4x + 23`

Divide through by 3

`y = (4)/(3)x + (23)/(3)`

Replace y with
`h^(-1)(x)`

`h^(-1)(x) = (4)/(3)x + (23)/(3)`

See attachment for graph