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Asymptote Calculator
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An asymptote is a straight line at a finite distance from the origin to which distance a tangent to a curve tends as the distance from the origin of the point of contact tends to infinity.

Asymptote Calculator (or Vertical Asymptote Calculator or Horizontal Asymptote Calculator) is an online tool which is used to calculate horizontal and vertical asymptotes. It takes a function and calculates horizontal and vertical asymptotes. Asymptotes calculator is sometimes known as kalkulator asymptot or oblique asymptote calculator or vertical asymptote finder.

Below is given a default function, click "Submit". To find vertical asymptotes, it sets the denominator to zero and solve it. To find horizontal asymptotes, first compare the degrees of numerator and denominator. In this case, degrees are same, therefore, it is solved by taking up the terms of same degree.

## Steps For Asymptote Calculator

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Step 1 :

Set the denominator equal to zero and solve for x in order to get the vertical Asymptote

Step 2 :

If the degrees of the numerator and the denominator are the same then this rational has a non-zero (non-x-axis) horizontal asymptote.

Solve the equation by taking up the terms of same degree. solve it for Y.

## Examples For Asymptote Calculator

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1. ### Find vertical and horizontal asymptotes for y = $\frac{x^{2} + 4x +1}{4x^{2} - 16}$?

Step 1 :

For vertical asymptote,

Set the denominator equal to zero and solve.
4x2 – 16 = 0

4x2 = 16

x2 = $\frac{16}{4}$
x = $\pm$ 2

Step 2 :

For horizontal asymptote

here degree is 2, by dividing the leading terms

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

y = $\frac{1}{4}$

Answer  :

Horizontal asymptote:  y = $\frac{1}{4}$
vertical asymptote: x = $\pm$ 2

2. ### Find vertical and horizontal asymptotes for y = $\frac{x +4}{x^{2} + 16}$?

Step 1 :

For vertical asymptote,

Set the denominator equal to zero and solve.
x2 + 16 = 0

x2 = -16

since the number is negative, we cant find the quare root for vertical Asymptote, so here he vertical Asymptote is none.

Step 2 :

now, here the degree is greater in the denominator than in the numerator so the y-values will be dragged down to the x-axis and the horizontal asymptote is therefore "y = 0"

Answer  :

horizontal asymptote:  y = 0
vertical asymptote: x = none

3. ### The actual minimum voltage was required to excite a nerve fiber is related to the time during which the current moves. As per the data collected by the college students, this relationship is modeled through the function:f(x) = $\frac{21}{0.153 + x}$ + 26.5Where x could be the time during which current flows (in milliseconds) and f(x) could be the minimum current was required to excite the nerve fiber (in milliseconds) Distinguish any vertical and horizontal asymptotes.

Step 1 :

f(x) = $\frac{21}{0.153 + x}$ + 26.5

For vertical asymptote, set the denominator of the function is equal to zero.

0.153 + x = 0

Or x = - 0.153

Step 2 :

For horizontal asymptote,

$\frac{21}{0.153 + x}$ : Degree of numerator  < degree of denominator

So x-axis is the horizontal asymptote of function, or y = 0

f(x) = 0 + 26.5 = 26.5.

Answer  :

26.5

4. ### For a university student with sensitive skin, the amount of time T the student can be exposed to the sun with minimal burning can be modeled byT = $\frac{0.47x + 43.8}{x}$ ; 0 < x $\leq$ 130Where x is the sensor scale reading. The sensor scale is based on the level of intensity of UVB rays. Find the amount of line a student with sensitive skin can be exposed to the sun with minimal burning when x = 20 and x = 50. If the model is true for x > 0, what would be the horizontal asymptote of this function. (Time is in hours)

Step 1 :

When x = 20, T = $\frac{0.47 \times 20 + 43.8}{20}$  = 2.66 hours

When x = 50, T = $\frac{0.47 \times 50 + 43.8}{50}$ = 1.346 hours

Step 2 :

And for horizontal asymptote, T = $\frac{coeff\ of x\of \ numerator}{ coeff\ of x\of \denominator}$

T = $\frac{0.47}{1}$

Horizontal asymptote is the line, T = 43.8.

Answer  :

T = 43.8.

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