# How would you find the equations of both the horizontal and vertical asymptotes for the following functions: y=(x)/(x-3)? y=(x+4)/(x^2-1)? y=(x+4)/(x^2-1)? y=(x^2-2x+1)/(x^2-3x-4)? y=(x^2-9)/(x^2-3x^2-18x)?

## In addition, how would you find the coordinates of any holes for these problems?

Aug 8, 2016

vertical asymptote at x = 3
horizontal asymptote at y = 1

#### Explanation:

For y $= \frac{x}{x - 3}$

The denominator of y cannot be zero as this is undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: x - 3 = 0 $\Rightarrow x = 3 \text{ is the asymptote}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , y \to c \text{ (a constant)}$

divide terms on numerator/denominator by x

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

as $x \to \pm \infty , y \to \frac{1}{1 - 0}$

$\Rightarrow y = 1 \text{ is the asymptote}$
graph{(x)/(x-3) [-10, 10, -5, 5]}

Aug 8, 2016

vertical asymptotes at x = ± 1
horizontal asymptote at y = 0

#### Explanation:

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

The explanation for vertical/horizontal asymptotes is as described above in previous question.

solve: x^2-1=0rArr(x-1)(x+1)=0rArrx=±1

$\Rightarrow x = - 1 \text{ and " x=1" are the asymptotes}$

To obtain horizontal asymptote divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

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

as $x \to \pm \infty , y \to \frac{0 + 0}{1 - 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(x+4)/(x^2-1) [-10, 10, -5, 5]}

Aug 11, 2016

For: $y = \frac{{x}^{2} - 2 x + 1}{{x}^{2} - 3 x - 4}$

#### Explanation:

Start by factoring both the numerator and the denominator:

y =((x - 1)(x - 1))/((x - 4)(x + 1)

Nothing can get canceled out, so we don't have any holes. Now for asymptotes. The vertical asymptotes can be found by setting the denominator to $0$ and solving for x.

${x}^{2} - 3 x - 4 = 0$

$\left(x - 4\right) \left(x + 1\right) = 0$

$x = 4 \mathmr{and} - 1$

So, the equations of the vertical asymptotes will be $x = 4$ and $x = - 1$.

For the horizontal asymptotes, we must examine the degree of the denominator relative to that of the numerator.

•The denominator is of degree $2$.
•The numerator is of degree $2$.

When the degrees are equal, the horizontal asymptote occurs at the ratio between the coefficients of the highest degree in the numerator and in the denominator.

Hence, $\text{asymptote"_"horizontal} = \frac{1 \times {x}^{2}}{1 \times {x}^{2}} = \frac{1}{1} = 1$

So, there will be a horizontal asymptote at $y = 1$.

In summary:

The graph of $y = \frac{{x}^{2} - 2 x + 1}{{x}^{2} - 3 x - 4}$ will have vertical asymptotes at $x = 4$ and $x = - 1$. It will have a horizontal asymptote at $y = 1$. The function will have no holes.

Bonus:

Here is the graph of the function. As you can see, the asymptotes are just where algebra showed them to be. Hopefully this helps!