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

We have a hole when

Let's factorise the numerator.

Therefore,

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

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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.

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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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