Question

Part 3 - Discussion/Explanation Question

Answer 1

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,

$\frac{1}{x - 5}$

Set the expression in the denominator equal to 0, because you can't divide by 0.

$x - 5 = 0$

$x = 5$

So the vertical asymptote is x=5.

Disclaimer if you see something like this

$\frac{(x - 5)(x + 3)}{(x - 5)}$

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

$\frac{1}{x}$

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

$a \times {x}^{0}$

For example,

$1 = 1 \times {x}^{0}$

$2 = 2 \times {x}^{0}$

And so on,

and

$x = {x}^{1}$

So our equation is basically

$\frac{1 \times {x}^{0} }{ {x}^{1} }$

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

$\frac{3 {x}^{2} }{ {x}^{2} + 1}$

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,

$x = \frac{3}{1}$

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.