Creating a Function with a Hole at x = -1 and Vertical Asymptote at x = 1

Hole when x = -1, Vertical Asymptote at x = 1

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In order to provide a detailed answer, we need to clarify the question a little bit. Is there a function provided or are we being asked to find a function that satisfies both of these conditions? I will assume that we are being asked to find a function that satisfies both of these conditions.

To satisfy the condition of having a hole at x = -1, the function must approach some finite value as x approaches -1, but the function must not be defined at x = -1. We can create a hole at x = -1 by first choosing some function that has a factor of (x + 1) in the numerator and denominator such as:

f(x) = (x + 1)/(x + 1)

This function is equal to 1 everywhere except at x = -1, where it is undefined. This creates a hole at x = -1 since the function approaches the value of 1 as x approaches -1 from both sides, but is not defined at x = -1.

To satisfy the condition of having a vertical asymptote at x = 1, the function must approach infinity or negative infinity as x approaches 1 from either the left or the right. This can be accomplished by adding a factor of (x – 1) to the denominator of the function we created earlier, giving us:

g(x) = (x + 1)/((x – 1)(x + 1))

The factor of (x – 1) in the denominator ensures that the function becomes unbounded as x approaches 1. However, the factor of (x + 1) in the numerator cancels out the hole at x = -1 that we created earlier. So the function g(x) has a vertical asymptote at x = 1, but is continuous everywhere else.

Therefore, the function g(x) = (x + 1)/((x – 1)(x + 1)) satisfies the conditions of having a hole at x = -1 and a vertical asymptote at x = 1.

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