How do you find the vertical, horizontal or slant asymptotes for #f(x)= (3x-5)/(x-6)#?

Question

Benjamin Achtenberg · Accepted Answer

vertical asymptote x = 6
horizontal asymptote y = 3

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero. solve : x - 6 = 0 → x = 6 is the asymptote Horizontal asymptotes occur as #lim_(xto+-oo) f(x) to 0 # divide terms on numerator/denominator by x #((3x)/x - 5/x)/(x/x - 6/x) = (3 - 5/x)/(1 - 6/x) # as # x to +- oo , 6/x" and " 5/x to 0 " and " y to 3/1 # #rArr y = 3 " is the asymptote " # Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes. graph{(3x-5)/(x-6) [-20, 20, -10, 10]}

Theodore Amidon · Answer

undefined To find the vertical asymptote, set the denominator equal to zero and solve for $x$.

For $x - 6 = 0$:
$x = 6$

So, the vertical asymptote is $x = 6$.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator of the function.

The degree of the numerator (1) is less than the degree of the denominator (1), so there is no horizontal asymptote.

To find the slant asymptote, if it exists, perform polynomial long division to divide the numerator by the denominator. In this case, $3x - 5$ is divided by $x - 6$.

$$ \frac{3x - 5}{x - 6} $$

Using polynomial long division, we find that the quotient is $3$ with a remainder of $13$.

So, the slant asymptote is $y = 3x + \frac{13}{x - 6}$.