# How do you find the Vertical, Horizontal, and Oblique Asymptote given #y = (x^2 - 5x + 6)/( x - 3)#?

This is the equation of the line

Let's factorise the numerator

So,

This is the equation of a line

graph{(x^2-5x+6)/(x-3) [-12.66, 12.65, -6.33, 6.33]}

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To find the vertical asymptote, set the denominator equal to zero and solve for ( x ). In this case, ( x - 3 = 0 ), so ( x = 3 ) is the vertical asymptote.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is ( y = 0 ). In this case, since the degree of the numerator (2) is less than the degree of the denominator (1), the horizontal asymptote is ( y = 0 ).

To find the oblique asymptote, divide the numerator by the denominator using polynomial long division or synthetic division. The oblique asymptote is the quotient obtained from this division. In this case, divide ( x^2 - 5x + 6 ) by ( x - 3 ) to get the oblique asymptote. The quotient is ( y = x - 2 ).

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To find the vertical asymptote, we set the denominator equal to zero and solve for x. In this case, the denominator is (x - 3), so setting it equal to zero gives us (x - 3 = 0), which yields (x = 3). Therefore, the vertical asymptote is (x = 3).

To find the horizontal asymptote, we examine the degrees of the numerator and denominator. Since the degree of the numerator (which is 2) is equal to the degree of the denominator (which is 1), there is no horizontal asymptote. However, we can use polynomial long division or synthetic division to find the slant or oblique asymptote.

To find the oblique asymptote, we divide the numerator by the denominator. After performing the division, the quotient will represent the equation of the oblique asymptote if the degree of the numerator is greater than the degree of the denominator. In this case, since the degree of the numerator (which is 2) is greater than the degree of the denominator (which is 1), we perform the division.

(x^2 - 5x + 6) divided by (x - 3) gives us (x - 2) with a remainder of 0. Therefore, the equation of the oblique asymptote is (y = x - 2).

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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.

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.

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.

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