I'm studying for a test and I came across this example problem and the oblique part is throwing me off. How do i go about solving this?
2 Answers
Look at the oblique and vertical asymptotes separately. An expression for a function with the vertical asymptote is $$y=\frac{1}{(x-3)(x-5)}$$ Since this goes to zero as x goes to infinity, it does not affect the oblique asymptote. Then you can consider the function $$y=\frac{1}{(x-3)(x-5)}+2x-3$$ This graph has all the features that you want. To convert to a rational form, you just need to find a common denominator to combine.
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Alright, so the denominator should obviously be something like $(x-5)(x-3)$, or $x^2-8x+15$.
To get the desired oblique asymptote, let the numerator equal the denominator times $2x-3$. However, make sure to add some random constant to the numerator, or else your rational equation will become a line. Thus, your equation should be something like this:
$$\frac{2x^3-19x^2+54x-45+C}{x^2-8x+15}$$
Where $C$ is some arbitrary nonzero constant.
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