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

Mar 19, 2018

See explanation

#### Explanation:

If you are dealing with questions at this level you know how to find the $x \mathmr{and} y$ intercepts
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Undefined at $x = 1$ as we have $y = \frac{- 2}{0}$ Thus we have vertical asymptotic behaviour
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$\textcolor{b l u e}{\text{Investigating } x \to 1}$

If we have $y = \frac{{\left({x}^{2} + \delta x\right)}^{2} - 3}{x + \delta x - 1}$ where $x = 1$

Then we have $y = \left(\text{negative")/("positive}\right) < 0$

x -> 0^+ }lim_(x->0^+)(x^2-3)/(x-1)->k where $k \to - \infty$

Conversely

If we have $y = \frac{{\left({x}^{2} - \delta x\right)}^{2} - 3}{x - \delta x - 1}$ where $x = 1$

Then we have $y = \left(\text{negative")/("negative}\right) > 0$

x-> 0^- }lim_(x->0^-)(x^2-3)/(x-1)->k where $k \to + \infty$
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$\textcolor{b l u e}{\text{Investigating } x \to + \infty}$

The temptation is to state the the const values become insignificant so we end up with $y = {x}^{2} / x \to y = x$
THIS IS WRONG

If we actually divide the denominator into the numerator we get

$y = x + \frac{x - 3}{x - 1}$

Now when we take limits we end up with $y = \infty + \frac{\infty}{\infty}$

Set $x = \infty$ gives $\textcolor{b l u e}{y = x + 1}$ as an oblique asymptote