Question

How do you find intercepts, extrema, points of inflections, asymptotes and graph `f(x)=(2x^2-5x+5)/(x-2)`?

Answer 1

Two asymptotes; two `x`-intercepts; one `y`-intercept; and two stationary points.

Explanation:

First of all do polynomial division to get your function in the nice form of:
`f(x) = 2x + 1 + 3/(x-2)`.

From this it is fairly clear where the asymptotes are, as `x` cannot equal `2` as `f(x)` would then be undefined. Similarly `f(x)` cannot equal `2x + 1`, as this would mean that `3/(x-2)` equals `0`, which it cannot as it implies `3=0`.

Then differentiate this function, applying the chain rule and product rules where necessary.

The derivative of `2x + 1` is of course `2`.

The derivative of `3/(x-2)` is done by letting `y=3(x-2)^-1`.

First apply the chain rule: say `t = x-2` and `w=t^-1`.
`:. (dt)/(dx) = 1` and `(dw)/(dt) = -t^-2`.
`:. (dw)/(dx) = -t^-2 = -(x-2)^-2`

The product rule isn't really necessary as it's pretty clear you will end up with:

`(dy)/(dx) = -3/(x-2)^2`.

`:. f'(x) = 2 -3/(x-2)^2`.

Letting `f'(x) = 0` next.

`:. 3 = 2(x-2)^2`

Simplify it down to arrive at `2x^2-8x+8 = 0`, which by the quadratic formula leaves us with: `x=(4-sqrt(6))/2` or `x=(sqrt(6)+4)/2`.

Solving for `f((4-sqrt(6))/2)` and `f((sqrt(6)+4)/2)` yields `5-sqrt(6)` and `2sqrt(6) + 5`, which are maximum and minimum respectively.

Finally calculate your intercepts, first by letting `f(x)=0`

`:. -3 = (2x+1)(x-2)`

`:. 2x^2-3x+1 =0`

`:. x = 1/2` or `x=1`.

Finally calculate `f(0)` which is simply `1-3/2`, therefore the `y`-intercept is where `y=-1/2`.