Home > Precalculus > Asymptotes

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

Answer 1

Madelyn Anderson

We have a hole when #x=2#

Let's factorise the numerator.

#x^2-5x+6=(x-3)(x-2)#

Therefore,

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

#=(x-2)#

We have a hole at #x=2#

#y=(x-2)# is the equation of a straight line.

graph{x-2 [-10, 10, -5, 5]}

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Liam Brennan

To find the vertical asymptote, set the denominator equal to zero and solve for ( x ). In this case, the vertical asymptote is ( x = 3 ).

To find the horizontal asymptote, compare the degrees of the numerator and denominator. Since the degree of the numerator (2) is greater than the degree of the denominator (1), there is no horizontal asymptote.

To find the slant (oblique) asymptote, perform polynomial long division of the numerator by the denominator. The quotient obtained represents the equation of the slant asymptote. After dividing ( x^2 - 5x + 6 ) by ( x - 3 ), the quotient is ( x - 2 ), which represents the equation of the slant asymptote.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.