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I need to write a rational function based on the following.

Only VA are x = 2 and x = -4, HA at y =3 , hole at (3,21) , and y –intercept is 3, and no x-
intercept.

VA is vertical asymptote and HA is horizontal asymptote. This is for precalc.

User Sharukh
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1 Answer

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Final answer:

To construct a rational function with the specified characteristics, we form the denominator to set the vertical asymptotes, include a canceled factor in the numerator for the hole, choose the leading coefficient for the horizontal asymptote, and adjust the numerator for the y-intercept without adding x-intercepts, resulting in the function f(x) = 3(x - 3)(x + 2)/(x - 2)(x + 4).

Step-by-step explanation:

To create a rational function with the given characteristics (only vertical asymptotes at x = 2 and x = -4, horizontal asymptote at y = 3, a hole at (3,21), and a y-intercept of 3, with no x-intercepts), you can start by forming the denominator to set the vertical asymptotes. The function must factor to (x - 2) and (x + 4) in the denominator:

f(x) = \frac{N(x)}{(x - 2)(x + 4)}

To ensure a hole at (3,21), the numerator must contain a factor of (x - 3), but it should be canceled out to not affect the x-intercepts:


f(x) = \frac{(x - 3)P(x)}{(x - 2)(x + 4)}

To have the horizontal asymptote at y = 3, the leading coefficients of the numerator and denominator should be equal when the function is simplified, meaning we need a leading coefficient of 3 in both the numerator and denominator:


f(x) = \frac{3(x - 3)(x + a)}{(x - 2)(x + 4)}

Given that the y-intercept is 3 (which implies that f(0) = 3), we must solve for 'a' by substituting x with 0:


f(0) = \frac{3(0 - 3)(0 + a)}{(0 - 2)(0 + 4)} = 3

Solve for 'a' and you get a = 2. Now you have enough information to write the function knowing that there are no x-intercepts (P(x) should not change the sign of the product). A possible function can be:


f(x) = \frac{3(x - 3)(x + 2)}{(x - 2)(x + 4)}

This function now satisfies all the criteria given.

User Filhit
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