- The domain is all real numbers except \(x=1\) and \(x=3\), written as \(\{x \mid x
eq 1, 3\}\).
- There is a hole at \((1, -1/2)\) and a vertical asymptote at \(x=3\).
- The function has no zeros, a y-intercept at \((0, -1/3)\), and a horizontal asymptote at \(y=0\).
- Final summary: \(\text{Domain: } (-\infty, 1) \cup (1, 3) \cup (3, \infty), \text{Zero(s): None, } y\text{-int: } (0, -1/3), \text{V.A.: } x=3, \text{Holes: } (1, -1/2), \text{H.A.: } y=0\).
Explanation
- Domain and Factoring
To analyze the function \(f(x) = \frac{x-1}{x^2-4x+3}\), we begin by examining the denominator to determine the domain and potential discontinuities. The quadratic expression \(x^2 - 4x + 3\) can be factored as \((x-1)(x-3)\). Thus, the function is defined as \(f(x) = \frac{x-1}{(x-1)(x-3)}\). The domain consists of all real numbers except those that make the denominator zero. Setting the factors equal to zero, we find \(x-1=0 \implies x=1\) and \(x-3=0 \implies x=3\). Therefore, the domain is \(\{x \in \mathbb{R} \mid x
eq 1, x
eq 3\}\), or in interval notation, \((-\infty, 1) \cup (1, 3) \cup (3, \infty)\).
- Simplification and Identifying Holes
We can simplify the expression for \(f(x)\) by canceling the common factor \((x-1)\) in the numerator and denominator, which is valid for all \(x\) in the domain (\(x
eq 1\)). This gives the simplified function:
\[f(x) = \frac{1}{x-3}, \text{ for } x
eq 1.\]
Because the factor \((x-1)\) was removed, there is a removable discontinuity (a hole) at \(x=1\). To find the \(y\)-coordinate of the hole, we evaluate the simplified function at \(x=1\): \(f(1) = \frac{1}{1-3} = -\frac{1}{2}\). Thus, there is a hole at \((1, -\frac{1}{2})\).
- Finding Vertical Asymptotes
A vertical asymptote occurs where the denominator of the simplified function is zero, provided the numerator is non-zero. Setting the denominator of the simplified function \(\frac{1}{x-3}\) to zero, we have \(x-3=0\), which gives \(x=3\). Since the numerator is a constant (1), it will never be zero at \(x=3\), so there is a vertical asymptote at \(x=3\).
- Zeros and y-intercept
To find the zeros of the function, we look for \(x\)-values in the domain where the function equals zero. For a rational function, this happens when the numerator of the simplified expression is zero. Since the simplified numerator is 1, which is never zero, there are no zeros for this function. For the \(y\)-intercept, we evaluate the function at \(x=0\):
\[f(0) = \frac{0-1}{0^2-4(0)+3} = \frac{-1}{3} = -\frac{1}{3}.\]
Thus, the \(y\)-intercept is at \((0, -\frac{1}{3})\).
- Horizontal Asymptote and Final Summary
The horizontal asymptote describes the behavior of the function as \(x\) approaches positive or negative infinity. We look at the limit of the simplified function:
\[\lim_{x \to \pm\infty} \frac{1}{x-3} = 0.\]
As \(x\) grows very large in either the positive or negative direction, the value of the fraction approaches 0. Therefore, the horizontal asymptote is the line \(y=0\).
Examples
In engineering, such rational functions often model the relationship between variables in a control system, where vertical asymptotes represent points of instability (poles) and horizontal asymptotes represent steady-state behavior as time or input tends to infinity. Understanding these features allows engineers to predict system behavior and avoid conditions that could cause system failure.