10. If f(x) =X-1/(x+1)(x-2)²Which is true about f?
To determine what is true about f(x), we need to evaluate its properties based on the given function f(x) = \(\frac{X-1}{(x+1)(x-2)^2}\).
1. Domain: The function f(x) is defined for all real numbers except for x = -1 and x = 2. This is because the denominator (x+1)(x-2)^2 cannot be zero.
2. Range: The range of f(x) is all real numbers except 0. This is because the numerator (x-1) can take any value, while the denominator cannot be zero. Therefore, f(x) can be any non-zero real number.
3. Asymptotes: The function f(x) has vertical asymptotes at x = -1 and x = 2. This is because the denominator (x+1)(x-2)^2 becomes zero at these points, resulting in the function going towards positive or negative infinity.
4. Intercepts: To find the x-intercept, we set f(x) equal to zero and solve for x. However, in this case, the numerator (x-1) can never be zero since it is not part of the equation. Therefore, f(x) has no x-intercepts.
5. The y-intercept occurs when x = 0: f(0) = \(\frac{0-1}{(0+1)(0-2)^2} = \frac{-1}{(1)(4)} = -\frac{1}{4}\). Therefore, the y-intercept is (0, -1/4).
6. End Behavior: As x approaches positive or negative infinity, f(x) approaches 0. This is because the degree of the numerator (1) is less than the degree of the denominator, and hence, the function tends towards zero.
In summary, the true statements about f(x) are:
- Its domain is all real numbers except for x = -1 and x = 2.
- Its range is all real numbers except 0.
- It has vertical asymptotes at x = -1 and x = 2.
- It has no x-intercepts.
- Its y-intercept is (0, -1/4).
- As x approaches positive or negative infinity, f(x) approaches 0.