How do you find the asymptotes for #h(x) = (x^3-8)/(x^2-5x+6)#?

Question

Serenity Johnson · Accepted Answer

Slant asymptote { #y=x+7.
Vertical asymptote : x = 3.

graph{((x^2+4x+4)/(x-3)-y)(y-x-7.9)(x-2+.01y)=0 [-45, 45, -20, 25]} The graph is a hyperbola #(y-x-7)(x-3)=15# having asymptotes given by #(y-x-7)(x-3)=0#, with a hole at (2, -16) #h =((x-2)/(x-2))((x^2+4x+4)/(x-3))#. Sans the hole at x = 2, #h=(x^2+4x+4)/(x-3)# #=x+7+25/(x-3)#. So, #y = quotient=x+7# gives the slant asymptote and #x-3 = 0# gives the vertical asymptote.

Lily Anderson · Answer

undefined To find the asymptotes of the function $ h(x) = \frac{x^3 - 8}{x^2 - 5x + 6} $, follow these steps:

1. Determine the vertical asymptotes by identifying the values of $ x $ for which the denominator becomes zero. Set the denominator equal to zero and solve for $ x $. These values will give you the vertical asymptotes.

$$ x^2 - 5x + 6 = 0 $$

$$ (x - 2)(x - 3) = 0 $$

$$ x = 2 \text{ or } x = 3 $$

So, the vertical asymptotes are $ x = 2 $ and $ x = 3 $.

2. To find the horizontal asymptote, compare the degrees of the numerator and the denominator of the rational function. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. If the degrees are equal, divide the leading coefficients to find the horizontal asymptote.

In this case, the degree of the numerator (3) is greater than the degree of the denominator (2), so there is no horizontal asymptote.

Therefore, the vertical asymptotes are $ x = 2 $ and $ x = 3 $, and there is no horizontal asymptote.