What are the horizontal and vertical asymptotes of r(x)=(6x+1)/(16x^2)+1?
1 answer:
Horiz. asy.: experiment with letting x grow larger and larger (without bound). Eventually the given expression will approach some limiting value.
If we continue to increase x in <span>(6x+1)/(16x^2), the fraction will approach 0. Try it!
If we continue to increase x in </span><span>(6x+1)/(16x^2) + 1, the expression will approach 0 + 1, or 1.
</span>
Which one did you mean?
Thus, the horiz. asy. would be y=0 or y=1.
Vertical asy.: Identify the x-value, if any, at which the denominator of this fraction = 0.
If you meant <span>(6x+1)/(16x^2) (without the +1), the fraction is undefined at x=0. Thus, the vert. asy. is the vertical line x=0.</span>
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