# How do you find the vertical, horizontal or slant asymptotes for #f(x)=( 2x+1)/(x-1)#?

vertical asymptote x = 1

horizontal asymptote y = 2

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

solve : x - 1 = 0 → x = 1 is the equation

If the degree of the numerator and denominator are equal , as in this case , both of degree 1 . Then the equation can be found by taking the ratio of leading coefficients.

Here is the graph of the function. graph{(2x+1)/(x-1) [-10, 10, -5, 5]}

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To find the vertical asymptote of the function ( f(x) = \frac{2x + 1}{x - 1} ), set the denominator equal to zero and solve for ( x ). In this case, ( x - 1 = 0 ), which gives ( x = 1 ) as the vertical asymptote.

To determine if there is a horizontal asymptote, compare the degrees of the numerator and denominator. Since the degree of the numerator (1) is less than the degree of the denominator (1), there is no horizontal asymptote.

To find any slant asymptotes, perform long division or polynomial division on the function. When you divide ( 2x + 1 ) by ( x - 1 ), you'll get a quotient of ( 2 ) and a remainder of ( 3 ). Therefore, there is a slant asymptote at ( y = 2x + 2 + \frac{3}{x - 1} ).

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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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