How do you find the asymptotes for #(x^2 - 5x + 6)/( x - 3)#?

Question

Avery Alexander · Accepted Answer

The asymptotes are #x=3# and #y=x-5#

An asymptote is defined as a line or curve that another line or curve approaches, but never meets. As such, we simply find the values of #x# and #y# that the y can never cross.

You only need to look at the denominator—which, as you already know, cannot equal 0—to determine the vertical asymptote because otherwise, the curve would not be defined.

So, the #x-3# in the denominator can never equal 0

#x-3!=0# #x!=3#

Therefore, #x=3# is the asymptote as #x# can never equal #3#, or else the denominator would become 0 and undefined.

To find the horizontal asymptote, divide every symbol/number in both the numerator and denominator by the highest #x# power in the denominator.

For example, in this equation, the highest #x# that can be found is #x^2#, but that is in the denominator. The highest power of #x# in the denominator, however, is simply #x#, and so that is what we will use.

So,

#(x^2/x-(5x)/x+6/x)/(x/x-3/x)# #(x-5+6/x)/(1-3/x)#

For all values that are any value over x, replace that with a 0 as, as #x# in the denominator gets infinitely larger, the overall number will get smaller and approach 0.

#y=x-5# is the horizontal/oblique asymptote.

graph{(x^2-5x+6)/x-3 [-3.095, 16.905, -4.44, 5.56]}

Eli Assouline · Answer

undefined To find the asymptotes for the function $\frac{x^2 - 5x + 6}{x - 3}$, we look for values of $x$ that make the denominator equal to zero, as these will result in vertical asymptotes. In this case, setting $x - 3 = 0$ gives $x = 3$. Therefore, there is a vertical asymptote at $x = 3$.

To find any horizontal asymptotes, we examine the behavior of the function as $x$ approaches positive or negative infinity. We do this by analyzing the degrees of the numerator and denominator polynomials. In this case, the degree of the numerator is 2, and the degree of the denominator is 1.

Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. Instead, there is a slant or oblique asymptote. We can find this by performing polynomial long division or synthetic division. After dividing $x^2 - 5x + 6$ by $x - 3$, we obtain $x - 2$ as the quotient. Therefore, the equation of the slant asymptote is $y = x - 2$.

In summary, the asymptotes for the given function are:
- Vertical asymptote: $x = 3$
- Slant asymptote: $y = x - 2$