What function has 2 vertical asymptotes at x=-3/2 and x=3/2 and a horizontal asymptote at 3/4? I know that the coefficients of the rational function are 3 and 4 but I don't know how to figure anything else out.
It should look like a corner in the II quadrant and IV quadrant and a -x^3 function in the middle.
vertical asymptotes: y = 1/((x+3/2)(x-3/2))
for a nonzero horizontal asymptote, the top and bottom must have the same degree, so
horizontal asymptote: y = 3x^2/(4(x^2 - 9/4))
If you want corners in QII and QIV, you'll have to adjust things a bit.
To determine the function that satisfies the given conditions, we can use the information about the vertical and horizontal asymptotes.
Let's start by considering the vertical asymptotes. A vertical asymptote occurs when the denominator of a rational function becomes zero. Therefore, the vertical asymptotes of the function will be located at the values of x that make the denominator of the function equal to zero.
In this case, we are given that the vertical asymptotes occur at x = -3/2 and x = 3/2. This means that the denominator of the function should be zero at these values.
To find the equation of the function, we need to determine the factors in the denominator that correspond to the vertical asymptotes. Since we have two vertical asymptotes, we can assume that the denominator of the function can be factored into two linear factors.
Let's set up the equation for the denominator as follows:
Denominator = (x - a)(x - b)
Now we can substitute the values of the vertical asymptotes (-3/2 and 3/2) into this equation:
(x - (-3/2))(x - (3/2)) = 0
Simplifying this expression gives us:
(x + 3/2)(x - 3/2) = 0
Expanding and simplifying further:
(x^2 - 9/4) = 0
Now we know that the denominator of the rational function is given by (x^2 - 9/4).
Next, let's consider the horizontal asymptote. A horizontal asymptote occurs when the degree of the numerator is equal to or less than the degree of the denominator.
Given that the coefficients of the rational function are 3 and 4, we can assume the function is in the form:
f(x) = (3x^n + 4) / (x^2 - 9/4)
Since there is a horizontal asymptote at 3/4, the degree of the numerator needs to be less than the degree of the denominator. Therefore, the degree of the numerator should be 0 (constant), and the degree of the denominator should be 2.
Now we can rewrite the function as:
f(x) = (C)/(x^2 - 9/4).
Where C is a constant.
So, the function that satisfies the given conditions is:
f(x) = C / (x^2 - 9/4).
To determine the value of C, we would need additional information, such as a specific point the function passes through or additional constraints. Without further information, we cannot determine the exact value of C or the specific function.