y=1/(x-1)(x+2)
The expression you've provided is a rational function given by:
\[ y = \frac{1}{(x-1)(x+2)} \]
Key Features of the Function
Domain: The function is undefined where the denominator is zero. Thus, we need to find the values of \(x\) that make the denominator zero:
\[ (x-1)(x+2) = 0 \]
This gives us the points:
- \(x - 1 = 0 \Rightarrow x = 1\)
- \(x + 2 = 0 \Rightarrow x = -2\)
Therefore, the domain of the function is all real numbers except \(x = 1\) and \(x = -2\). In interval notation, the domain is:
\[ (-\infty, -2) \cup (-2, 1) \cup (1, \infty) \]
Vertical Asymptotes: Since the function is undefined at \(x = 1\) and \(x = -2\), these values correspond to vertical asymptotes. As \(x\) approaches 1 or -2, the function will approach \(\pm \infty\).
Horizontal Asymptotes: As \(x\) approaches \(\pm \infty\), the degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator. Therefore, the horizontal asymptote is:
\[ y = 0 \]
Intercepts:
y-intercept: Set \(x = 0\):
\[ y = \frac{1}{(0-1)(0+2)} = \frac{1}{(-1)(2)} = -\frac{1}{2} \]
x-intercept: Set \(y = 0\). A rational function can only be zero if the numerator is zero, but here, the numerator is 1, which is never zero. Thus, there are no x-intercepts.
Summary of Important Properties
- Domain: \( (-\infty, -2) \cup (-2, 1) \cup (1, \infty) \)
- Vertical Asymptotes: \(x = 1\), \(x = -2\)
- Horizontal Asymptote: \(y = 0\)
- y-intercept: \((0, -\frac{1}{2})\)
- x-intercept: None
Graphing the Function
When graphing this function, consider the following points and behaviors:
- The function approaches the vertical asymptotes \((x = 1\) and \(x = -2)\) and will have branches in the intervals defined by the vertical asymptotes.
- The graph will approach the horizontal asymptote \(y = 0\) as \(x\) goes to \(\infty\) or \(-\infty\).
You can sketch the graph using a variety of graphing tools or software to visualize the behavior more clearly.
