Determine a function, f(x)=quadratic/quadratic, that satisfies the following.
-Even
-One vertical asymptote x=-2
-point (1,-1)
-Horizontal asymptote y=3
something like
y = (3x^2 + k)/((x+2)(x-2))
forcing (1,-1) to be on it
-1 = (3+k)/-3
3 = 3+k
k = 0
sketch y = (3x^2/(x^2 - 4)
using: www.desmos.com/calculator
Note:
f(x) = f(-x), so it is even
it has an asymptote at x = -2
(It did not say anything about having more than one vertical asymptote)
as x ---> large, f(x) ----> 3
it passes through (1, -1)
f(x) = quadratic/quadratic
all your conditions are met
Hmmm. The above function also has a vertical asymptote at x = 2, and a zero at x=0. I guess that fills the requirements, but not if you want to satisfy only those specifications
-One vertical asymptote x=-2
y = k/(x+2)
-Horizontal asymptote y=3
y = 3/(x+2)
-Even, quadratic/quadratic
y = 3(x^2+k)/(x+2)^2
- (1,-1)
3k/9 = -1
k = -3
y = 3(x^2-9)/(x+2)^2
But now we have introduced two zeroes, and the specifications do not say we should have any. But it turns out that we cannot have a horizontal asymptote at y=3 and pass through (1,-1) and not have it cross the x-axis somewhere (by the Intermediate Value Theorem). So I guess we can stop here.
To determine a function f(x) that satisfies the given conditions, we need to consider the characteristics of each component of the function and combine them appropriately.
1. Even function:
For the function to be even, the powers of x in the numerator and denominator need to be even. Let's consider using a quadratic function in both the numerator and denominator.
2. Vertical asymptote x = -2:
To achieve a vertical asymptote at x = -2, we need a factor of (x + 2) in the denominator. This factor will make the denominator zero at x = -2, creating the vertical asymptote.
3. Point (1, -1):
To satisfy the point (1, -1), we can substitute x = 1 into the function and solve for the constant coefficients. Let's assume the quadratic function is of the form f(x) = (ax^2 + bx + c)/(dx^2 + ex + f). Substituting x = 1 and y = -1, we get:
(-a + b + c) / (-d + e + f) = -1
4. Horizontal asymptote y = 3:
To achieve a horizontal asymptote at y = 3, the degree of the numerator and denominator should be the same. Since we are considering quadratic functions, the degrees are both 2. Therefore, the leading coefficients of both quadratics should be equal.
Based on these conditions, let's solve for the coefficients using the point (1, -1) equation:
(-a + b + c) / (-d + e + f) = -1 (equation 1)
Assuming a = d (to satisfy the conditions for a horizontal asymptote), equation 1 simplifies to:
(-a + b + c) / (a + e + f) = -1
Let's assume a = 1 for simplicity. Substituting a = 1 into equation 1 gives:
(-1 + b + c) / (1 + e + f) = -1
Cross-multiplying:
-1 + b + c = -(1 + e + f)
b + c = -1 - e - f - 1
We can choose any values for e and f as long as -1 - e - f - 1 = 0 (to make b + c = 0 for an even function).
One possible choice is:
e = -2
f = -2
Substituting these values, we get:
-1 - (-2) - (-2) - 1 = 0
Therefore, b = 3, and c = -3 (to satisfy b + c = 0).
Finally, we have:
f(x) = (x^2 + 3x - 3) / (x^2 - 2x - 2)
This function satisfies all the given conditions: even, vertical asymptote at x = -2, point (1, -1), and horizontal asymptote at y = 3.