Question

How do you graph `y =(x^2-3)/(x-1)`?

Answer 1

See explanation

Explanation:

If you are dealing with questions at this level you know how to find the `x and y` intercepts
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Undefined at `x=1` as we have `y=(-2)/0` Thus we have vertical asymptotic behaviour
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`color(blue)("Investigating "x->1)`

If we have `y=((x^2+deltax)^2-3)/(x+deltax-1)` where `x=1`

Then we have `y=("negative")/("positive") <0`

x -> 0^+ `}lim_(x->0^+)(x^2-3)/(x-1)->k` where `k->-oo`

Conversely

If we have `y=((x^2-deltax)^2-3)/(x-deltax-1)` where `x=1`

Then we have `y=("negative")/("negative") >0`

x-> 0^- `}lim_(x->0^-)(x^2-3)/(x-1)->k` where `k->+oo`
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`color(blue)("Investigating "x->+oo)`

The temptation is to state the the const values become insignificant so we end up with `y=x^2/x->y=x`
THIS IS WRONG

If we actually divide the denominator into the numerator we get

`y=x+(x-3)/(x-1)`

Now when we take limits we end up with `y=oo+(oo)/(oo)`

Set `x=oo` gives `color(blue)(y=x+1)` as an oblique asymptote