How do you find the slant asymptote of $\frac{x^{2} + 3 x + 2}{x - 2}$ $(x^2+3x+2)/(x-2)$ ?

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How do you find the slant asymptote of $\frac{x^{2} + 3 x + 2}{x - 2}$ $(x^2+3x+2)/(x-2)$ ?

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Answer 1 · 2015-12-27T00:42:29

Divide to find the quotient polynomial $y = x + 5$ $y = x+5$ , which is the slant asymptote (otherwise known as the oblique asymptote).

Explanation:

Long divide or do something like this to separate out the quotient $(x + 5)$ $(x+5)$ and remainder $12$ $12$ :

$\frac{x^{2} + 3 x + 2}{x - 2}$ $(x^2+3x+2)/(x-2)$

$= \frac{x^{2} - 2 x + 5 x - 10 + 12}{x - 2}$ $=(x^2-2x+5x-10+12)/(x-2)$

$= \frac{(x^{2} - 2 x) + (5 x - 10) + 12}{x - 2}$ $=((x^2-2x)+(5x-10)+12)/(x-2)$

$= \frac{x (x - 2) + 5 (x - 2) + 12}{x - 2}$ $=(x(x-2)+5(x-2)+12)/(x-2)$

$= \frac{(x + 5) (x - 2) + 12}{x - 2}$ $=((x+5)(x-2)+12)/(x-2)$

$= x + 5 + \frac{12}{x - 2}$ $=x+5 + 12/(x-2)$

Then as $x \to \pm \infty$ $x->+-oo$ the term $\frac{12}{x - 2} \to 0$ $12/(x-2) -> 0$

So the slant asymptote is $y = x + 5$ $y = x+5$

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How do you find the slant asymptote of $\frac{x^{2} + 3 x + 2}{x - 2}$ $(x^2+3x+2)/(x-2)$ ?

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Explanation:

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