Can someone check my homework?

i).

We are given `f(x)=(x-2)^2(x-5)` and `x=1 ,x=5`

First find the y coordinates by plugging x into `f(x)`

`x=1`

`f(1)=((1)-2)^2((1)-5)=-4`

Coordinates: `color(blue)((1,-4)`

`x=5`

`f(5)=((5)-2)^2((5)-5)=0`

Coordinates: `color(blue)((5,0)`

These are the endpoints of the line segment.

Gradient is:

`"change in y"/("change in x")=(y_2-y_1)/(x_2-x_1)`

`:.`

`(0-(-4))/(5-1)=4/4=1`

Gradient is 1

ii)

The coordinate of the midpoint are given by:

`((x_1+x_2)/2,(y_1+y_2)/2)`

`((1+5)/2,(0-4)/2)->color(blue)((3,-2)`

If this point lies of `f(x)` then:

`f(3)=-2`

`f(3)=((3)-2)^2((3)-5)=-2`

This lies on `f(x)`

iii)

If the gradients are are equal ie `1`, the simplest way to find these is to differentiate `f(x)`:

expanding `f(x)`:

`f(x)=x^3-9x^2+24x-20`

`dy/dx( x^3-9x^2+24x-20)=3x^2-18x+24`

Since this is a gradient function and we need the gradient to be `1`:

`3x^2-18x+24=1`

`3x^2-18x+23=0`

We solve this for `x`:

Using quadratic formula give:

`x=(9+2sqrt(3))/3`, `x=(9-2sqrt(3))/3`

These are the values of `x` for which the tangent is `1`:

This is confirmed by the graph:

I don't know if you are supposed to use calculus methods to answer iii. This is the way I would solve it.

You seem to have the wrong results for i and ii. I don't know how you got `f(1)=24`, this would have made ii wrong as well.

Hope this helps.