How do you find the slant asymptote of $f(x) = (3x^2 + 2x - 5)/(x - 4)$ ?

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Answer 1 · 2018-07-10T13:46:37

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When you notice that the degree of your numerator is greater than your denominator, then it is likely that you will need to use long division.

$f(x)=(3x^2+2x-5)/(x-4)=((x-4)(3x+14)+51)/(x-4)=(3x+14)+51/(x-4)$

To find your slant or oblique asymptote, you are finding what happens when x approaches infinity.

When x approaches infinity in the above equation, $51/(x-4)$ will approach zero (try putting big numbers into your calculator)

Hence, $f(x)$ becomes $3x+14+0=3x+14$
Therefore, your slant asymptote is $y=3x+14$

How do you find the slant asymptote of f(x) = (3x^2 + 2x - 5)/(x - 4)f(x) = (3x^2 + 2x - 5)/(x - 4)?