Construct a rational function that has a vertical asymptote at x = 3 and a removable discontinuity at x = -2. Explain how you determined your answer.
To construct a rational function with a vertical asymptote at x = 3 and a removable discontinuity at x = -2, we need to consider the factors in the numerator and denominator that would create these characteristics.
The general form of a rational function is f(x) = (ax^n + bx^(n-1) + ... + c)/(dx^m + ex^(m-1) + ... + f). We need to choose the values of the coefficients a, b, c, d, e, and f such that the given conditions are satisfied.
Firstly, we want a vertical asymptote at x = 3. This means we want a factor of (x - 3) in the denominator, as this would create a vertical line at x = 3. So, the denominator will include the factor (x - 3).
Secondly, we want a removable discontinuity at x = -2. This means we want both the numerator and denominator to have a factor of (x - (-2)) = (x + 2). However, we want this factor to cancel out in the simplified form of the rational function, as a removable discontinuity means that the function can be simplified by removing that factor.
To achieve this, we can include this factor (x + 2) in both the numerator and denominator. However, we want it to cancel out when simplified. We can choose the coefficient in the numerator, let's say a, to be the negative of the coefficient in the denominator, let's say d. This way, when we simplify, the factor (x + 2) cancels out and the function becomes removably discontinuous at x = -2.
Putting these pieces together, a possible rational function that satisfies the given conditions is:
f(x) = a(x + 2) / (d(x - 3)(x + 2))
The choice of the values of coefficients a and d can be further adjusted to obtain a specific shape for the graph and additional characteristics like horizontal asymptotes or x-intercepts.