horizontal asymptote?

Question

can sum1 explain to me how to get the horizontal asymptote for (x² - 3x - 7)/(x+3) without a graphing calculator

Answer 1

Take the limit as x approaches infinity. If the limit L exists, then y = L is a horizontal asymptote. Also take the limit as x approaches negative infinity to (possibly) get another horizontal asymptote.

Answer 2

There is NO horizontal asymptote...if there is a larger power of x (or whatever the variable is) in the numerator than the denominator, there is not a horizontal asymptote. In this case, the highest power in the numerator is x^2, while the highest power in the denominator is x^1 or just x. Therefore, a horizontal asymptote does not exist.

If you switched the numerator and denominator in your problem, the horizontal asymptote would be 0 because the highest power is in the denominator.

If the highest powers in the numerator and denominator are equal, you just divide them. This is the basic rule you can follow for all examples...the horizontal asymptote does not exist in this case because it would be 1/0 (the coefficients of the highest power, in this case, the power of 2). Obviously, 1/0 DNE since you cannot have 0 in the denominator of a number. And if the highest power is on the bottom, the division will be 0/(coefficient of highest power in denominator) which will equal zero no matter what the number is. That's how you do it.

Answer 3

This doesn't have a horizontal asymptote, since the degree in the numerator (2) is higher than the degree of the denominator (1)

It has a slant asymptote since the top degree is exactly one higher than the bottom. Thus, you have to do the long division to find the slant asymptote.

x^2 -3x - 7 / x + 3 divides out to x - 6 with remainder of 11
disregard the remainder and the slant asymptote is y = x -6

Answer 4

vertical asymptote is where y is undefined let y = f(x)= -2x+2/x thus y = -2x^2 + 2 = 2( -x^2 -1 ) = 2 (1+x) (1-x) thus y is undefined at x = -1 and x = 1 those are vertical asymptotes horizontal asymptotes is value of y when x approaches infinity : y = -2x^2 + 2 when x goes to infinity +2 and coffecient of x^2 becomes negligible thus y= -x^2 that is y approaches negative infinity : )

Answer 5

you must divide (x^2 -3x - 7) by (x+3). It is not going to be exactly horizontal but instead will be slanted since the degree of the X on the numerator is one bigger than the degree on of the X in the denominator. IF you get a remainder for the answer when you divide. Just disregard it and what ever your main answer is, is the equation to the asymptote.

Answer 6

y = (x² - 3x - 7)/(x + 3) has no horizontal asymptote

It has a vertical asymptote at x = -3

If you divide x² - 3x - 7 by x + 3 you get

x - 6 + 11/(x + 3)

y = x - 6 is an asymptote

Answer 7

If you had a horizontal asymptote I'll tell you how lol.

The degree of the numerator is larger than the degree of the denominator meaning there is no horizontal asymptote.

Answer 8

horizontal asymptotes are found by taking the limits of the function as x goes to plus and minus infinity where the function is defined (domain)

In this case

the limit goes to + infinity

no horizontal asymptote

Answer 9

You take the highest power function(s) (here it would be x^2)
and divide by them. For example, for the function (5x^2+17)/(3x^2+7x+9), the horizontal asymptote would be 5/3. In your case, it would be 0, since 1/0 is zero.
(In the denominator, there is 0x^2)

Answer 10

you need to take the degreeof the first term of each part
so in this one its x^2 and x
the degrees are 2 and 1.. so the asymptote will be 0