Question

Graph each of the following functions: $$ y=\frac{x^{2}-1}{x-1} \quad y=\frac{x^{3}-1}{x-1} y=\frac{x^{4}-1}{x-1} \quad y=\frac{x^{5}-1}{x-1} $$ Is \(x=1\) a vertical asymptote? Why? What happens for \(x=1 ?\) What do you conjecture about the graph of \(y=\frac{x^{n}-1}{x-1}, n \geq 1\) an integer, for \(x=1 ?\)

Short answer

\(x=1\) is a removable discontinuity, not a vertical asymptote. There is a hole at \(x=1\). For \(y=\frac{x^n-1}{x-1}\), there is always a hole at \(x=1\).

Step-by-step solution

- Simplify the Functions

First, simplify each of the given functions by factoring the numerator and then canceling the common factor with the denominator. For example, for the function \(y=\frac{x^2-1}{x-1}\), we can factor the numerator as \((x-1)(x+1)\), which simplifies to \(y=x+1\) with a restriction. Similarly, apply this process to the other functions. 1. \(y=\frac{x^2-1}{x-1} = x+1\) for \( x eq 1\) 2. \(y=\frac{x^3-1}{x-1} = x^2 + x + 1\) for \(x eq 1\) 3. \(y=\frac{x^4-1}{x-1} = x^3 + x^2 + x + 1\) for \( x eq 1\) 4. \(y=\frac{x^5-1}{x-1} = x^4 + x^3 + x^2 + x + 1\) for \( x eq 1\)

- Identify Restrictions

Note that for each simplified function, \(x\) cannot be equal to 1 because the original function had a division by zero at that point, creating a discontinuity.

- Determine Vertical Asymptote

Check if \(x=1\) is a vertical asymptote by examining the limits. For each function, since the simplified form is a polynomial and does not have a discontinuity at any point other than \(x=1\), \(x=1\) is actually a removable discontinuity, not a vertical asymptote.

- Evaluate Functions at \(x=1\)

Evaluate the original functions at \(x=1\). For each original function, \(y\) is undefined at \(x=1\) due to division by zero. Thus, there is a hole in the graph at \(x=1\).

- General Conjecture for \(y=\frac{x^n-1}{x-1}\)

For any integer \(n \geq 1\), the graph of \(y=\frac{x^n-1}{x-1}\) has a hole at \(x=1\) because the denominator \(x-1\) cancels out with one of the factors in the numerator, leading to a simplified polynomial function with a removable discontinuity at \(x=1\).

Key concepts

Factorization

Factorization is the process of breaking down an expression into a product of simpler expressions. In the context of rational functions, it helps us simplify complex functions.

For example, to factor the numerator of the rational function \( y = \frac{x^2 - 1}{x - 1} \), we recognize it as a difference of squares: \( x^2 - 1 = (x - 1)(x + 1) \).

After factorization, the function simplifies to \( y = x + 1 \) for all \( x eq 1 \). This simplification process makes it easier to understand the behavior of the function near the problematic points, such as where the denominator is zero.

Vertical Asymptote

A vertical asymptote is a line that a graph approaches but never touches or crosses. It occurs when the output value of a function approaches infinity as the input gets close to a certain value, making the function undefined at that point.

For the functions we examined, such as \( y = \frac{x^2 - 1}{x - 1} \), \( x = 1 \) initially appears to be a candidate for a vertical asymptote because it makes the denominator zero. However, after simplifying the function, we see that \( x = 1 \) is merely a removable discontinuity and not a vertical asymptote.

Discontinuity

Discontinuities occur when a function is not continuous at a certain point. There are various types of discontinuities, including removable, jump, and infinite discontinuities.

In our step-by-step solution, we find that \( x = 1 \) is a removable discontinuity for the simplified rational functions. This means that although the function is undefined at \( x = 1 \), we can 'remove' this problem by simplifying the function. For example, after simplifying \( y = \frac{x^2 - 1}{x - 1} \) to \( y = x + 1 \), the only issue at \( x = 1 \) is a hole in the graph rather than an asymptote or a jump.

Polynomial Functions

Polynomial functions are mathematical expressions consisting of variables raised to non-negative integer powers and coefficients. They are smooth and continuous, without breaks or holes.

In our exercise, we dealt with polynomial functions that were initially presented as rational functions. For example, \( y = \frac{x^3 - 1}{x - 1} \) simplifies to the polynomial \( y = x^2 + x + 1 \) after factoring and canceling out common terms. This simplified form is easier to graph and understand, especially when identifying points of discontinuity and behavior near those points.