Math Exploration: Creating Equations with a Removable Discontinuity and Vertical Asymptote

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

To find the equation of a function that satisfies the given conditions, we can make use of the hole at x = -1 and the vertical asymptote at x = 1

To find the equation of a function that satisfies the given conditions, we can make use of the hole at x = -1 and the vertical asymptote at x = 1.

Let’s start by addressing the hole at x = -1. When there is a hole in the graph of a function, it means that there is a removable discontinuity at that point. This typically occurs when the function has a factor that cancels out, resulting in an undefined value.

To find the equation of the function that has a hole at x = -1, we need to cancel out a factor that causes the discontinuity. One way to do this is by introducing the factor (x + 1) in both the numerator and the denominator, and then simplifying.

Let’s denote the function as f(x) and rewrite it as:

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

We canceled out the (x + 1) factor to remove the discontinuity at x = -1. Now, the function g(x) represents the behavior of the function f(x) at the hole.

Next, let’s address the vertical asymptote at x = 1. A vertical asymptote occurs when the function approaches infinity or negative infinity as x approaches a specific value.

To create a vertical asymptote at x = 1, we can introduce a factor (x – 1) in the denominator without canceling it out. This will ensure that the function approaches infinity or negative infinity as x approaches 1. Let’s modify our function to include this factor:

f(x) = (x + 1)(g(x))/(x + 1)(x – 1)

Now, our function has the desired properties. However, we need to ensure that g(x) does not cancel out any factors, as it represents the behavior of the function at the hole. So, g(x) should not have any common factors with (x – 1).

An example of a function that satisfies these conditions could be:

f(x) = (x + 1)(x^2 – 1) / (x + 1)(x – 1)

Here, g(x) is represented by (x^2 – 1). Notice that this function does not cancel out the factor (x – 1) in the denominator.

This equation represents a function that has a removable discontinuity (hole) at x = -1 and a vertical asymptote at x = 1.

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