How do you find vertical, horizontal and oblique asymptotes for #f(x) = (3x + 5) /( x - 2)#?

Question

Christian Antunez · Accepted Answer

vertical asymptote at x = 2
horizontal asymptote at y = 3

The denominator of f(x) cannot be zero as this would make f(x) 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 - 2 = 0 #rArrx=2" is the asymptote"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by x f(x)#=((3x)/x+5/x)/(x/x-2/x)=(3+5/x)/(1-2/x)# as #xto+-oo,f(x)to(3+0)/(1-0)# #rArry=3" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1 ) Hence there are no oblique asymptotes. graph{(3x+5)/(x-2) [-20, 20, -10, 10]}

Benjamin Achtenberg · Answer

undefined To find the vertical asymptote of the function $ f(x) = \frac{3x + 5}{x - 2} $, set the denominator equal to zero and solve for $ x $. The vertical asymptote occurs where the function is undefined, which happens when the denominator is zero:

$$ x - 2 = 0 $$
$$ x = 2 $$

So, the vertical asymptote is $ x = 2 $.

To find horizontal asymptotes, determine the limit of the function as $ x $ approaches positive or negative infinity. If the limit exists, it represents the horizontal asymptote.

$$ \lim_{x \to \infty} \frac{3x + 5}{x - 2} = 3 $$
$$ \lim_{x \to -\infty} \frac{3x + 5}{x - 2} = 3 $$

Since the limit exists and is finite, the horizontal asymptote is $ y = 3 $.

For oblique asymptotes, divide the numerator by the denominator using polynomial long division or synthetic division. In this case, dividing $ 3x + 5 $ by $ x - 2 $ gives:

$$ 3 $$

So, the oblique asymptote is $ y = 3 $.