When the denominator equals zero, there is a vertical asymptoe
When the numerator becomes constant as x approaches very large, there is a horizontal asymptoe.
For instance, in the second
q(x)=(x-1)/x= 1-1/x as x gets very large, it becomes 1-0 or 1
In the first,
f(x)=x/(x-1)= 1/(1-1/x) (multiplied numerator and denominator by 1/x)
f(x)= 1/(1-0) when x is very large, or f(x)=1
x = 1 for the first one
x = 0 for the second
yes, for the vertical asy.
so what about this one g(x)=1.5^x
No vertical, no horizontal.
how can you tell it has neither
As Mr. Bob mentioned:
When the denominator becomes zero when x=c and c is finite, there is a vertical asymptote.
A horizontal asymptote is typically identified by the fact that lim f(x) approaches a constant value as x->∞ or -> -&infin.
In the case of:
there is no denominator that makes g(x) infinite when x is finite, so no vertical asymptote.
g(x) becomes infinite when x->+∞, so no horizontal asymptote on the right. But on the left..., as x->-∞, g(x) approaches zero, so what do you think?
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