Given the function
f(x)=x^3-2x^2-5x+6/x^2+3x+2
a) Determine algebraically the equation for any asymptotes (HA, VA and or OA)
b)State the domain of the function
c)State the behaviour of the function at each of the asymptotes
since the denominator is (x+1)(x+2) you'd expect to see vertical asymptotes at x= -1 and -2
But the numerator is (x-1)(x+2)(x-3) so at everywhere except x = -1,-2
f(x) = (x-1)(x+2)(x-3) / (x+1)(x+2) = (x-2)(x-3) / (x+1)
= x-5 + 8/(x+1)
So, there is a slant asymptote of y=x-5
So, the only vertical asymptote is at x = -2. There is a hole at x = -1, since f(-1) = 0/0 is undefined.
There is no horizontal asymptote since the numerator has greater degree than the denominator.
For x ≠ -1, -2
f(x) =
sorry - lost track of where I was on the screen. But, the info is there/
its ok, i just didn't know how to verify if there was a horizontal asymptote and I didn't know that x=-1 is hole, thank you for the help!
To determine the asymptotes of the function, follow these steps:
Step 1: Find the vertical asymptotes (VA):
- Vertical asymptotes occur when the denominator of the rational function is equal to zero.
So, set the denominator, x^2+3x+2, to zero and solve for x.
x^2+3x+2 = 0
To solve this quadratic equation, you can factor it:
(x+1)(x+2) = 0
So, the solutions are x = -1 and x = -2.
Therefore, the vertical asymptotes are x = -1 and x = -2.
Step 2: Find the horizontal asymptotes (HA):
- Horizontal asymptotes occur when the degree of the numerator is less than or equal to the degree of the denominator.
In this case, the degree of the numerator is 3, and the degree of the denominator is 2.
Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
Therefore, the equation for the asymptotes is:
Vertical asymptotes: x = -1 and x = -2
No horizontal asymptote.
To determine the domain of the function, note that the function is a rational function, and the denominator cannot be equal to zero.
So, the domain of the function is all real numbers except for the values of x that make the denominator, x^2+3x+2, equal to zero.
In this case, the denominator factors to (x+1)(x+2), so the domain of the function is all real numbers except -1 and -2.
The behavior of the function at each asymptote can be determined by evaluating the function as x approaches the values of the asymptotes:
- At x = -1 and x = -2 (vertical asymptotes), you can evaluate the function by plugging in nearby values and observing the trend of the output.
- As x approaches -1 or -2 from the left or right side, the function's values will approach positive or negative infinity.
Note: You can also use a graphing calculator or software to visualize the function and determine the behavior at the asymptotes.