# How do you find the vertical, horizontal and slant asymptotes of: #y = (1-5x)/(1+2x)#?

vertical asymptote at

horizontal asymptote at

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

Horizontal asymptotes occur as

divide terms on numerator/denominator by x

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes. graph{(1-5x)/(1+2x) [-10, 10, -5, 5]}

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To find the vertical asymptotes, set the denominator equal to zero and solve for ( x ). For the function ( y = \frac{1 - 5x}{1 + 2x} ), the denominator ( 1 + 2x ) equals zero when ( x = -\frac{1}{2} ). Thus, there is a vertical asymptote at ( x = -\frac{1}{2} ).

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is ( y = 0 ). If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote. In this case, since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is ( y = 0 ).

To find the slant asymptote, perform polynomial long division on the numerator by the denominator. In this case, divide ( 1 - 5x ) by ( 1 + 2x ). The quotient is ( -3 ), and the remainder is ( 11 ). Therefore, the slant asymptote is ( y = -3x + 11 ).

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