Find the verticle, horizontal, and oblique asymtopes if any.
#1 (x^2 +6x+5)/2x^2 +7x+5
y= (1/2) and x= -5/2 and -1
#2 (8x^2 +26x-7)/4x-1
y=2x+7 and x=1/4
#3. (x^4-16)/(x^2-2x)
x= 0 and 2
y= x^2 +2x+4
thanks!
#1, I think you meant:
(x^2 + 6x+5)/(2x^2 + 7x + 5)
then
y = (x+1)(x+5)/((x+1)(2x+5))
= (x+5)/(2x+5)
horizontal: y = 1/2 , you had that
vertical x = -5/2
a "hole" at x = -1 , not a vertical asymptote
#2, again, you need brackets for the denominator, the way it stands , only the 4x is divided.
the rest is correct.
#3. The asymptote is actually the parabola
y = x^2 + 2x + 4 , so it is a "curved asymptote"
To find the vertical, horizontal, and oblique asymptotes, you need to analyze the behavior of the function as x approaches infinity or negative infinity.
1. For the function (x^2 + 6x + 5) / (2x^2 + 7x + 5):
- Vertical asymptotes: To find the vertical asymptotes, you need to determine the values of x that make the denominator equal to zero. So, set 2x^2 + 7x + 5 = 0 and solve for x. In this case, there are no real values of x that make the denominator equal to zero, which means there are no vertical asymptotes for this function.
- Horizontal asymptote: To find the horizontal asymptote, divide the leading terms of the numerator and denominator. In this case, the leading terms are both x^2. Therefore, the horizontal asymptote is y = 1/2.
- Oblique asymptote: If the degree of the numerator is greater than the degree of the denominator by exactly one, then there is an oblique asymptote. So, divide the numerator by the denominator using polynomial long division or synthetic division. The quotient will give you the equation of the oblique asymptote. However, in this case, the degrees of the numerator and denominator are the same; hence there is no oblique asymptote for this function.
2. For the function (8x^2 + 26x - 7) / (4x - 1):
- Vertical asymptotes: Set the denominator, 4x - 1, equal to zero and solve for x. In this case, x = 1/4 is the value that makes the denominator equal to zero. Therefore, there is a vertical asymptote at x = 1/4.
- Horizontal asymptote: Divide the leading terms of the numerator and denominator, which are 8x^2 and 4x, respectively. The result is 2x. Hence, the horizontal asymptote is y = 2x.
- Oblique asymptote: Since the degree of the numerator is equal to the degree of the denominator, there is no oblique asymptote for this function.
3. For the function (x^4 - 16) / (x^2 - 2x):
- Vertical asymptotes: Set the denominator, x^2 - 2x, equal to zero and solve for x. In this case, x = 0 and x = 2 are the values that make the denominator equal to zero. Therefore, there are vertical asymptotes at x = 0 and x = 2.
- Horizontal asymptote: Divide the leading terms of the numerator and denominator, which are x^4 and x^2, respectively. The result is x^2. Hence, the horizontal asymptote is y = x^2.
- Oblique asymptote: Since the degree of the numerator is greater than the degree of the denominator by two, there is no oblique asymptote for this function.
Please note that these are general guidelines, and the interpretation of asymptotes may vary based on the context and mathematical conventions.