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

1 Answer
Tony B
Mar 19, 2018

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
Tony B

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