Write an equation of a rational function that satisfies all of these conditions below. Show how each condition is satisfied in order to get full marks.
- y-intercept of -5
- x-intercepts at x=-1 and x=5
- Hole at x=-2
- Horizontal asymptote at y=6
- No vertical asymptotes
Hole at x=-2 ||| y = (x+2)/(x+2)
x-intercepts at x=-1 and x=5 ||| y = (x+2)(x+1)(x-5)/(x+2)
Horizontal asymptote at y=6 ||| y = 6(x+2)(x+1)(x-5) / (x+2)(x^2+1)
now, y(0) = -30
so, divide by -6 at x=0 and you get
y = 6(x+2)(x+1)(x-5) / (x+2)(x^2-6)
see the graph at
https://www.wolframalpha.com/input/?i=%286+%28x%2B2%29%28x%2B1%29%28x-5%29%29+%2F+%28%28x%2B2%29%28x%5E2-6%29%29+
To find an equation of a rational function that satisfies the given conditions, we can start by considering the information provided and work step by step.
1. The y-intercept of -5:
The y-intercept occurs when x=0 in the equation of the function. So, to satisfy this condition, let's assume the equation has the form:
f(x) = (x-a)(x+1)(x-5)/g(x)
where 'a' is a constant and g(x) is some polynomial expression.
Plugging in x=0, we get:
f(0) = (0-a)(0+1)(0-5)/g(0)
-5 = -5a/ g(0)
To have a y-intercept of -5, we can let a = 1, and we assume g(0) = 1.
So, the function becomes:
f(x) = (x-1)(x+1)(x-5)/g(x)
2. X-intercepts at x = -1 and x = 5:
The x-intercepts occur when f(x) = 0. Plugging in these values for x into the equation, we get:
(x+1)(x-1)(x-5)/g(x) = 0
To have x-intercepts at -1 and 5, we can set the numerator (x+1)(x-1)(x-5) = 0.
(x+1)(x-1)(x-5) = 0
(x+1)(x^2-5x-x+5) = 0
(x+1)(x^2-6x+5) = 0
(x+1)(x-1)(x-5) = 0
So, the function becomes:
f(x) = (x+1)(x-1)(x-5)/g(x)
3. Hole at x = -2:
A hole occurs when there is a common factor in the numerator and denominator that cancels out. To create a hole at x = -2, we need a common factor of (x+2) in both the numerator and denominator.
Let's assume we introduce a factor of (x+2) in both the numerator and denominator:
f(x) = (x+1)(x-1)(x-5)(x+2)/(x+2)g(x)
This creates a hole at x = -2 since (x+2)/(x+2) simplifies to 1, canceling out the common factor.
4. Horizontal asymptote at y = 6:
For a rational function to have a horizontal asymptote at y = a, the degree of the numerator should be less than or equal to the degree of the denominator.
To have a horizontal asymptote at y = 6, we want the numerator to have a lower degree than the denominator. Thus, we have:
f(x) = (x+1)(x-1)(x-5)(x+2)/(x+2)(x-b)
Here, the denominator degree is greater due to the factor (x-b), and a reasonable choice would be to let b = 0.
So, the function becomes:
f(x) = (x+1)(x-1)(x-5)(x+2)/(x+2)x
5. No vertical asymptotes:
To avoid vertical asymptotes, we need to ensure the denominator doesn't equal zero for any x-values. In our function, we have (x+2) in the denominator, but this factor is also present in the numerator, canceling it out. Therefore, there are no vertical asymptotes.
Combining all the conditions, the equation of the rational function that satisfies all the given conditions is:
f(x) = (x+1)(x-1)(x-5)(x+2)/(x+2)x
Each condition has been demonstrated step by step, showing how it is satisfied in the final equation.