Understanding the Vertical Asymptote at x=4 and the Hole at x=-3 in a Function: Explained with Examples

Vertical Asymptote at x= 4 and a hole when x = -3

To understand why there is a vertical asymptote at x=4 and a hole at x=-3, let’s first define what they mean

To understand why there is a vertical asymptote at x=4 and a hole at x=-3, let’s first define what they mean.

A vertical asymptote is a vertical line on a graph where the function approaches either positive infinity or negative infinity as x approaches a certain value. In this case, the vertical asymptote is at x=4, which means that as x gets closer and closer to 4 from either the left or the right side, the function will approach either positive or negative infinity.

On the other hand, a hole is a point on a graph where a specific x-value is excluded from the function due to simplifications or cancellations in the equation. In this case, there is a hole at x=-3, which means that our function is undefined at that point.

Now, let’s go step by step to determine why these conditions occur in the function:

First, let’s consider the vertical asymptote at x=4. To have a vertical asymptote at a certain value, the function must be approaching infinity or negative infinity as x gets closer to that value. In other words, as x approaches 4, the function must become very large in either the positive or negative direction.

To understand this further, we need to look at the behavior of the function as x approaches 4. Let’s assume our function is f(x). We can express this behavior by examining the limit of the function as x approaches 4, which can be denoted as:

lim (x -> 4) f(x)

If this limit exists and is finite, there would not be a vertical asymptote at x=4. However, if the limit either does not exist or approaches infinity or negative infinity, we have a vertical asymptote at x=4.

Let’s consider an example to illustrate this. Suppose we have the function f(x) = (x^2 – 16)/(x-4). If we evaluate the limit as x approaches 4, we get:

lim (x -> 4) [ (x^2 – 16)/(x-4) ]

If we directly substitute 4 into the function, we get an indeterminate form of 0/0. To overcome this, we can factorize the function:

lim (x -> 4) [ (x+4)(x-4)/(x-4) ]

Now, we can cancel out the common factor (x-4) which results in:

lim (x -> 4) [ x+4 ]

Finally, if we substitute 4 into the simplified function, we get:

4 + 4 = 8

Since the limit of the function as x approaches 4 exists and is finite (8), there is no vertical asymptote at x=4 in this case.

Moving on to the hole at x=-3, we need to find a function that has a factor (x+3) which can be canceled out by another factor in the denominator. This cancellation results in the function not being defined at x=-3.

As an example, let’s take the function f(x) = (x^2 – 9)/(x+3). If we evaluate the function for x=-3, we get:

f(-3) = [(-3)^2 – 9]/(-3+3)
= [9 – 9]/0

The denominator becomes 0, which is undefined. This indicates that the function is not defined at x=-3, and thus, we have a hole at that point.

In summary, there is a vertical asymptote at x=4 if the limit of the function as x approaches 4 either does not exist or approaches infinity or negative infinity. And there is a hole at x=-3 if there is a common factor in the numerator and denominator of the function, which can be canceled out resulting in an undefined value at x=-3.

More Answers: