Rational Functions
y = (x^2+7x-4)/(x-2)
Since the degrees of both the numerator and denominator are the same, I would use the leading coefficient, and both the rational functions start with 1, so would the horizontal asymptote be 1?
Also don't get how to find the horizontal asymptote of the following :
y = (2+(x+3)/(x-9)*(5-(x^2+6x-3)/(x^2-8x)
y = ((x+5)/(x^2+6x-9))*((3x^2+2x+8)\((x^2+7x-9))
How can you say "the degrees of both the numerator and denominator are the same", when the numerator clearly shows an x^2 term and the denominator only an x term ?
A division of x^2 + 7x - 4 by (x-2)
= x+9 + 14/(x-2)
so as x ---->large we converge on y = x+9
so there is no horizontal asymptote, but rather an oblique one of y = x+9
There is a vertical asymptote of x = 2
for y = (2+(x+3)/(x-9)*(5-(x^2+6x-3)/(x^2-8x)
both (x+3)/x-9) and (x^2 + 6x - 3)/(x^2-8) approach a value of 1 for each as x becomes large
so as x --->∞ , you are left with
y = (2 + 1)*(5-1)
= 12
so we have a horizonatal asymptote of y = 12
the last one:
as x --->∞, I see it as
y = (1/x)*(3/1) = 3/x
which zero infinity as x --->∞
So the horizontal asymptote is y = 0
sorry that was the wrong question I meant the degrees of these one were both the same:
y = x^2+5x-6/x-3x^2+9
the leading coefficient is 1, so the vertical asymptote would be one?
for the last two you did how do you know the x is becoming large like that whole infinity thing?
To find the horizontal asymptote of a rational function, you need to determine the behavior of the function as x approaches positive infinity or negative infinity. Here's how you can find the horizontal asymptote for the given rational functions:
1) y = (x^2+7x-4)/(x-2)
To find the horizontal asymptote, you compare the degrees of the numerator and the denominator. In this case, the degree of the numerator (2) is greater than the degree of the denominator (1).
If the degree of the numerator is greater than the degree of the denominator by 1, there is no horizontal asymptote. Instead, there will be a slant asymptote (oblique asymptote).
To find the slant asymptote, perform long division of the numerator by the denominator. The quotient will be the equation of the slant asymptote.
Performing long division on (x^2+7x-4)/(x-2), the quotient will be x + 9 and the remainder will be 14.
Therefore, the slant asymptote is y = x + 9.
2) y = (2+(x+3)/(x-9)*(5-(x^2+6x-3)/(x^2-8x)
To find the horizontal asymptote, again compare the degrees of the numerator and the denominator. In this case, the degree of the numerator is equal to the degree of the denominator (both are 3).
When the degrees of the numerator and denominator are the same, the horizontal asymptote can be found by comparing the coefficients of the leading terms of the numerator and denominator.
For simplicity, let's rewrite the rational function as:
y = [(2(x - 9) + (x + 3)(5 - (x^2 + 6x - 3)))] / [(x^2 - 8x)(x - 9)]
Simplifying further:
y = (2x - 18 + (x + 3)(5 - x^2 - 6x + 3)) / (x^3 - 17x^2 + 72x - 72)
Expanding and simplifying the numerator:
y = (2x - 18 + 5x - x^3 - 6x^2 + 3x + 15 - 3x^3 - 18x^2 + 9x) / (x^3 - 17x^2 + 72x - 72)
Combining like terms:
y = (-4x^3 - 24x^2 + 29x - 3) / (x^3 - 17x^2 + 72x - 72)
The leading terms of the numerator and denominator are -4x^3 and x^3, respectively.
Therefore, the horizontal asymptote is y = -4x^3 / x^3, which simplifies to y = -4.
So, for the second rational function, the horizontal asymptote is y = -4.