What is how to find vertical asymptotes?

Here's how to find vertical asymptotes:

A vertical asymptote is a vertical line that a function approaches but never touches. They typically occur where a rational function is undefined, meaning the denominator is zero.

Here's the general procedure:

1. Simplify the Function: Reduce the function to its simplest form by canceling out any common factors between the numerator and denominator. This step is crucial because a common factor that cancels out represents a hole (removable discontinuity), not a vertical asymptote.

2. Find the Zeros of the Denominator: Set the denominator of the simplified function equal to zero and solve for x. The values of x you obtain are potential locations of vertical asymptotes.

3. Check for Existence: For each potential vertical asymptote at x = a, verify that the limit of the function as x approaches a from the left and right is either positive or negative infinity.

   • lim (x→a^-) f(x) = ±∞
   • lim (x→a⁺) f(x) = ±∞

   If these limits exist, then x = a is a vertical asymptote. If the limit exists and is a finite number, then the function has a hole or is defined at x = a.

Important Considerations:

• Rational Functions: Vertical asymptotes are very common in rational functions, which are functions expressed as a ratio of two polynomials.
• Logarithmic Functions: Logarithmic functions often have vertical asymptotes at the point where the argument of the logarithm is zero.
• Tangent Functions: The tangent function and other trigonometric functions can also have vertical asymptotes. For example, tan(x) has vertical asymptotes at x = π/2 + nπ, where n is an integer.
• Piecewise Functions: Be mindful when dealing with piecewise functions, as the definition changes based on intervals, meaning that you should check points of discontinuity.

Example:

Consider the function f(x) = (x² - 4) / (x - 2).

1. Simplify: f(x) = (x + 2)(x - 2) / (x - 2) = x + 2, for x ≠ 2.

2. Denominator Zeros: The original denominator (x - 2) is zero at x = 2.

3. Check for Existence: Since the factor (x-2) cancels out, there is no vertical asymptote. Instead, there is a hole at x = 2. If the original function was f(x) = 1/(x-2), there would be a vertical asymptote at x = 2.