Determine the value of a so that $f(x) = \frac{x^2+ ax + 5}{x+1} $has a slant asymptote y = x + 3
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You want the following limit to exist and equal zero: $$\lim_{x \to \infty} [f(x) - (x+3)] = 0$$ Solve for $a$. You want the numerator to be a constant.
$$f(x) = \frac{x^2+ ax + 5}{x+1} = (x+3) + \frac{b}{x+1} \, \text{($b$ constant)}$$
Last edited by Archie; Apr 11th 2018 at 12:19 PM.
So a=4 and b=2
Yes, that is correct. For x not equal to 3, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}= x+ 3$.
Originally Posted by HallsofIvy Yes, that is correct. For x not equal to 3, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}= x+ 3$. Small typo: For x not equal to 1, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}+\frac{2}{x+1}= x+ 3 + \frac{2}{x+1}$.
Last edited by SlipEternal; Apr 12th 2018 at 06:21 AM.
Do you mean "for X not equal to -1"?
Originally Posted by Archie Do you mean "for X not equal to -1"? Lol, good point. Fixed.
Now it says "X not equal to 1", not "X not equal to -1"
Originally Posted by SlipEternal Small typo: For x not equal to 1, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}+\frac{2}{x+1}= x+ 3 + \frac{2}{x+1}$. Yes. Thank you and Archie.
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