How do you find the vertical, horizontal and slant asymptotes of: #y=(2x )/ (x-5)#?
vertical asymptote
horizontal asymptote
Horizontal asymptotes occur as
Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both degree 1 ) Hence there are no slant asymptotes. graph{(2x)/(x-5) [-20, 20, -10, 10]}
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To find the vertical asymptotes of the function , set the denominator equal to zero and solve for . These values will give you the vertical asymptotes of the function. To find the horizontal asymptote, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at . If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. To find the slant asymptote, perform polynomial long division of the numerator by the denominator and the quotient represents 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.
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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