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The horizontal asymptote is 0.

Q: What is the horizontal asymptote of 5 divided by the quantity of x minus 6?

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y = 1. When the degree of your numerator is the same with the degree of your denominator, then y = the ratio of the leading coefficients of the numerator and denominator is the horizontal asymptote.

-1

(4 - x)/3

The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3

(-9) / 3 = -3 . Minus nine divided by three is equal to minus three

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y = 1. When the degree of your numerator is the same with the degree of your denominator, then y = the ratio of the leading coefficients of the numerator and denominator is the horizontal asymptote.

-1

(4 - x)/3

The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3

irdk

An oblique asymptote is another way of saying "slant asymptote."When the degree of the numerator is one greater than the denominator, an equation has a slant asymptote. You divide the numerator by the denominator, and get a value. Sometimes, the division pops out a remainder, but ignore that, and take the answer minus the remainder. Make your "adapted answer" equal to yand that is your asymptote equation. To graph the equation, plug values.

Solve? 2 is where the right piece of this function crosses the X axis, but the vertical asymptote is important here.(X - 3)/(X + 2)divide both terms, top and bottom by X- 3/X divided by 2/Xsame as- 3/X * X/2= - 3/2========the vertical asymptote

÷/÷ = +

Plus

4 divided by 9 minus 1 divided 12?

12.600000000000001

-3