An asymptote is a straight line Which is is either parallel to x-axis or y-axis or cut the axis . The asymtote acts as a boundary for the graph of a function. When a function has an asymptote(horizontal/vertical or slant and not all functions have them) the function gets closer and closer to the asymptote as the input value to the function approaches either a specific value a or positive or negative infinity. The rational functions most likely to have asymptotes.

An asymptote is a line that shows that the curve approaches but does not cross the X and Y axis. The denominator q(x) of the rational function gives us the vertical asymptote. It can be found by finding the roots of the denominator or q(x). While finding the vertical asymptote we will ignore the numerator.

2) Set the denominator equal to zero and solve the for the given variable that (if any) gives you the vertical asymptotes,everything else is the domain.

1)Find the vertical asymptote for $f(x)=\frac{5x}{x-1}$

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

Here the denominator is x-1.

Set the denominator equal to zero and find the value of x.

So, x- 1 = 0

$\Rightarrow $ x= 1

So the vertical asymptote is at x= 1.

2) Find the vertical asymptote for $f(x)=\frac{4x}{x^2-1}$.

$y= \frac{4x}{x^2-1}$

Here the denominator is $x^{2}$-1.

Set the denominator equal to zero and find the value of x.

So, $x^{2}$-1 = 0

$\Rightarrow x^{2}$= 1

$\Rightarrow x =\pm$ 1

So the vertical asymptote is at x= 1 and x = -1.

3) Find the vertical asymptote for $f(x)=\frac{4x^2}{x^2+4}$

$y= \frac{4x^2}{x^2+4}$

Here the denominator is $x^{2}$ + 4.

Set the denominator equal to zero and find the value of x.

So, $x^{2}$ + 4 = 0

$\Rightarrow x^{2}$= -4

$\Rightarrow x =\pm \sqrt{-4}$

Since $\sqrt{-4}$ is not a real number, the graph will have no vertical asymptotes.

4) From the graph, find the vertical asymptote.

1) $f(x)= 2- \frac{2}{x-3}$

2) $f(x)= \frac{x^3}{2x^2-6}$

3)$f(x)= \frac{2}{x}$

4) $f(x)= \frac{x^2-5x+4}{x^2-4}$

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