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$$y = \frac{3}{8x - 3} $$

The y-intercept is $-1$ and the vertical asymptote is $x = \frac{3}{8}$ but what would be the horizontal asymptote and the x-intercept in this case?

I am asking this as the numerator is not a linear function (it is just a constant 3 so I do not know what the horizontal asymptote or root would be). I follow the procedure below:

$$y = \frac{ax + b}{cx + d}$$

root at $x = \frac{-b}{a}$ intercept at $y = \frac{b}{d} $

vertical asymptote at $x = \frac{-d}{c} $ horizontal asymptote at $y = \frac{a}{c}$

When finding the root, you get $0 = \frac{3}{8x - 3}$ and then $0 = 3$ which is not true, therefore this must mean the curve does not cut the x-axis?

The horizontal asymptote (using what I posted above) would be $ y = \frac{a}{c}$ which is $\frac{0}{8}$ hence the horizontal asymptote is $ y = 0$?

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For the horizontal asypmtotes, you want to see what happens when $x$ gets larger and larger (towards $\infty$), or when it gets smaller and smaller (towards $-\infty$).

Note that $\frac{3}{8x-3}$ gets closer and closer to zero for larger values of $x$ (because the degree of $8x-3$ is greater than the degree of the constant $3$). That means that the horizontal asymptote to the right is 0. The same goes for the left side.

Note that, for the method you included, there is no $ax$-term in the numerator, hence $a=0$, and so you get the horizontal asymptotes at $y=\frac{0}{c}=0$, just as it should be.

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When the degree of the denominator is greater than the degree of the numerator, limits at $\infty$ and at $-\infty$ are both $0$. And yes, if you divide $3$ by any number (as in the function) the result can never be $0$, so there is no $x$-intercept.

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  • $\begingroup$ I have not studied limits, ever. So that statement confuses me somewhat (i.e. I do not fully understand it). $\endgroup$ – Noah Apr 15 '14 at 13:19

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