Find the horizontal asymptote as ×-->8and describe what this mean in practical terms f(×)=150×+120/0.05×+1;the number of base ,f(×),after x months in a lake that was stocked with 120 bass
If you divide top and bottom by x, you have
f(x) = (150 + 120/x) / (.05 + 1/x)
As x gets huge, the fractions vanish, and
f(x) → 150/.05 = 3000
So, as time passes, the fish population stabilizes at 3000
so is this one of these answers; a. the number of bass, fx, after y months in a lake that was stocked by 130 bass. b. the number of bass, fx after x months in a lake that was stocked with 120 bass. c. the number of bass fx after x months in a lake was stocked with 140 bass. d. the number of bass fx after x months in a lake was stocked with 150 bass. i had b as my answer
To find the horizontal asymptote of the given function f(x), which represents the number of bass in a lake after x months, as x approaches 8, follow these steps:
Step 1: Rewrite the given function:
f(x) = (150x + 120)/(0.05x + 1)
Step 2: Determine the degree of the numerator and the denominator:
The numerator's degree is 1 (highest power of x is 1) and the denominator's degree is also 1.
Step 3: Divide the leading term of the numerator by the leading term of the denominator:
150x / 0.05x = 3000
Step 4: The horizontal asymptote is the ratio obtained in step 3:
The horizontal asymptote is y = 3000.
In practical terms, this means that as the number of months (x) increases, the number of bass in the lake (f(x)) will approach 3000. In other words, the population of bass will tend to stabilize around 3000 over time.
To find the horizontal asymptote of the given function f(x), which represents the number of bass after x months in a lake, we need to evaluate the limit of the function as x approaches infinity (x → ∞), and as x approaches negative infinity (x → -∞).
Given the function:
f(x) = (150x + 120) / (0.05x + 1)
1. As x approaches infinity (x → ∞):
Evaluate the limit of f(x) as x approaches infinity by dividing the leading terms of the numerator and denominator by the highest power of x:
f(x) = (150x/x + 120/x) / (0.05x/x + 1/x)
Simplifying:
f(x) = (150 + 120/x) / (0.05 + 1/x)
As x approaches infinity, the term 120/x and 1/x become negligible (approaching zero), and we are left with:
f(x) ≈ 150 / 0.05
Simplifying further:
f(x) ≈ 3000
Therefore, as x approaches infinity, the function f(x) approaches the value 3000. This means that as time goes on (as x gets larger and larger), the number of bass in the lake will approach 3000.
2. As x approaches negative infinity (x → -∞):
Following the same process as above, we divide the leading terms by the highest power of x:
f(x) = (150x/x + 120/x) / (0.05x/x + 1/x)
Simplifying:
f(x) = (150 + 120/x) / (0.05 + 1/x)
As x approaches negative infinity, the terms 120/x and 1/x become negligible, approaching zero. We are left with:
f(x) ≈ 150 / 0.05
f(x) ≈ 3000
Therefore, as x approaches negative infinity, the function f(x) also approaches the value 3000. This means that as time goes backward (as x gets more negative), the number of bass in the lake will still approach 3000.
In practical terms, the horizontal asymptote at f(x) = 3000 represents the long-term equilibrium point for the number of bass in the lake. Regardless of the initial stock of 120 bass, as time progresses or goes backward, the number of bass in the lake will eventually stabilize or converge to approximately 3000.