Question

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?

Answer 1

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

Explanation:

For y `=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 `rArrx=3" is the asymptote"`

Horizontal asymptotes occur as

`lim_(xto+-oo),ytoc" (a constant)"`

divide terms on numerator/denominator by x

`(x/x)/(x/x-3/x)=1/(1-3/x)`

as `xto+-oo,yto1/(1-0)`

`rArry=1" is the asymptote"`
graph{(x)/(x-3) [-10, 10, -5, 5]}

Answer 2

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

Explanation:

For `y=(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`

`rArrx=-1" 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`

`(x/x^2+4/x^2)/(x^2/x^2-1/x^2)=(1/x+4/x^2)/(1-1/x^2)`

as `xto+-oo,yto(0+0)/(1-0)`

`rArry=0" is the asymptote"`
graph{(x+4)/(x^2-1) [-10, 10, -5, 5]}

Answer 3

For: `y = (x^2 - 2x + 1)/(x^2 - 3x - 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 - 3x - 4 = 0`

`(x - 4)(x + 1) = 0`

`x = 4 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, `"asymptote"_"horizontal" = (1 xx x^2)/(1 xx x^2) = 1/1 = 1`

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

In summary:

The graph of `y = (x^2 - 2x + 1)/(x^2 - 3x - 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!