Create a function that has a graph with the following features:
Two vertical asymptotes: x=1, x=3. One horizontal: y = -1. And x intercepts= -2, 4
how about
f(x) = ((x+2)(x-4))/((3-x)(x-1)) ?
Yes, the function you provided, f(x) = ((x+2)(x-4))/((3-x)(x-1)), satisfies the given conditions. Let's analyze each feature:
1. Two vertical asymptotes: x = 1, x = 3.
If we simplify the denominator, we have (3 - x)(x - 1). As x approaches 1, the denominator becomes 0, indicating a vertical asymptote at x = 1. Similarly, as x approaches 3, the denominator also becomes 0, indicating another vertical asymptote at x = 3.
2. One horizontal asymptote: y = -1.
To determine the horizontal asymptote, we examine the behavior of the function as x approaches positive or negative infinity. If we simplify the function, we see that both the numerator and denominator are polynomials of degree 1. Therefore, the ratio of the leading terms is the determining factor. In this case, the leading term in the numerator is x^2, and in the denominator, it is -x^2. Thus, as x approaches infinity or negative infinity, the ratio of the leading terms is -1, indicating a horizontal asymptote at y = -1.
3. x-intercepts: -2, 4.
To find the x-intercepts, we set the numerator equal to zero since the x-coordinate of a point on the graph is where the function value is zero. In this case, if we set (x+2)(x-4) = 0, we find two solutions: x = -2 and x = 4, which confirms the given x-intercepts.
Therefore, the function f(x) = ((x+2)(x-4))/((3-x)(x-1)) satisfies all the given conditions.