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Question: Give an example of a rational function that has vertical asymptote $x=3$ now give an example of one that has vertical asymptote $x=3$ and horizontal asymptote $y=2$. Now give an example of a rational function with vertical asymptotes $x=1$ and $x=-1$, horizontal asymptote $y=0$ and x-intercept 4.

My solution: $(a) \frac{1}{(x-3)}$. This is because when we find vertical asymptote(s) of a function, we find out the value where the denominator is $0$ because then the equation will be of a vertical line for its slope will be undefined.

$(b) \frac{2x}{(x-3)}$. Same reasoning for vertical asymptote, but for horizontal asymptote, when the degree of the denominator and the numerator is the same, we divide the coefficient of the leading term in the numerator with that in the denominator, in this case $\frac{2}{1} = 2$

$(c) \frac{(x-4)}{(x-1)(x+1)}$. Same reasoning for vertical asymptote. Horizontal asymptote will be $y=0$ as the degree of the numerator is less than that of the denominator and x-intercept will be 4 as to get intercept, we have to make $y$, that is, $f(x)=0$ and hence, make the numerator 0. So, in this case; to get x-intercept 4, we use $(x-4)$ in the numerator so that $(x-4)=0 \implies x=4$.

Are my solutions correct of have I missed anything, concept-wise or even with the calculations?

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  • $\begingroup$ All right, you work is fine :) $\endgroup$ – Emilio Novati Jun 1 '16 at 12:55
  • $\begingroup$ @EmilioNovati Thanks! :) Could you also put that as an answer so that I can accept it? $\endgroup$ – MathEnthusiast Jun 1 '16 at 13:02
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    $\begingroup$ I agree with @EmilioNovati. Note that since the example in (a) has horizontal asymptote $y = 0$, so we can modify it as $\frac{1}{x - 3} + 2$ to give another answer to (b). $\endgroup$ – Travis Willse Jun 1 '16 at 13:06
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Your work is correct. Note that your solutions are the ''more simple'' rational functions that satisfies the requests. Obviously you can find infinitely many other rational functions that do the same, but have some other property.

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