Given the function:
f(x)=(x^2+6x+5)/(x^2+3x−10)
Find the
Domain:
Vertical asymptote(s):
Hole(s):
Horizontal asymptote (if exists):
f(x)=(x^2+6x+5)/(x^2+3x−10)
= (x+1)(x+5(/((x+5)(x-2) )
= (x+1)/(x-2)
domain: any real value of x , x ≠ -5, 2
vertical asymptote:
x = 2
hole:
x = -5
horizontal asymptote:
y = 1
see:
http://www.wolframalpha.com/input/?i=plot+y+%3D+%28x%5E2%2B6x%2B5%29%2F%28x%5E2%2B3x−10%29+
notice the graph cannot show the hole at x = -5
To find the domain of a function, we need to determine the values of x for which the function is defined.
In this case, the function f(x) is a rational function where the denominator is x^2 + 3x − 10. To find the domain, we need to make sure that the denominator is non-zero.
The denominator can be factored as (x + 5)(x − 2). Therefore, the function is undefined when the denominator equals zero.
Setting the denominator equal to zero, we get:
x + 5 = 0 or x - 2 = 0
Solving these equations, we find:
x = -5 or x = 2
Therefore, the function is undefined at x = -5 and x = 2.
The domain of the function is the set of all real numbers except for -5 and 2.
Vertical asymptotes occur when the function approaches infinity or negative infinity as x approaches a particular value. In this case, we can determine vertical asymptotes by observing the behavior of the function as x approaches infinity and negative infinity.
As x approaches infinity (i.e., x → ∞), the terms with lower powers become insignificant compared to the highest power terms, x^2, in both the numerator and denominator. Thus, the function approaches the value of:
f(x) = (x^2)/(x^2) = 1
As x approaches negative infinity (i.e., x → -∞), the same logic applies, and the function again approaches 1.
Therefore, the horizontal asymptote of the function is y = 1.
To find the holes in the function, we need to check if any common factors can be canceled out from the numerator and denominator.
In this case, we can factor the numerator as (x + 1)(x + 5) and factor the denominator as (x + 5)(x - 2). We notice that (x + 5) appears in both the numerator and denominator.
Canceling out the common factor (x + 5), we get:
f(x) = (x + 1)/(x - 2)
Now, the function has no common factors in the numerator and denominator, so there are no holes.
To recap:
- The domain is all real numbers except for -5 and 2.
- The vertical asymptotes occur at x = -5 and x = 2.
- There are no holes in the function.
- The horizontal asymptote exists at y = 1.