Step-by-step explanation:
Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.
For example,
[tex] \frac{1}{x - 5} [/tex]
Set the expression in the denominator equal to 0, because you can't divide by 0.
[tex]x - 5 = 0[/tex]
[tex]x = 5[/tex]
So the vertical asymptote is x=5.
Disclaimer if you see something like this
[tex] \frac{(x - 5)(x + 3)}{(x - 5)} [/tex]
x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.
Horizontal:
If we have a function like this
[tex] \frac{1}{x} [/tex]
We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.
Remember a constant can be represent by
[tex]a \times {x}^{0} [/tex]
For example,
[tex]1 = 1 \times {x}^{0} [/tex]
[tex]2 = 2 \times {x}^{0} [/tex]
And so on,
and
[tex]x = {x}^{1} [/tex]
So our equation is basically
[tex] \frac{1 \times {x}^{0} }{ {x}^{1} } [/tex]
Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.
So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.
Consider the function
[tex] \frac{3 {x}^{2} }{ {x}^{2} + 1} [/tex]
As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator
So in this case,
[tex]x = \frac{3}{1} [/tex]
Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.
