Find an equation of a rational function that satisfies the following conditions:
• Vertical asymptotes: x = −3
• Horizontal asymptote: y=3/2
• x -intercept: 5
• Hole at x =2
• Vertical asymptotes: x = −3
y = 1/(x+3)
• Hole at x =2
y = (x-2)/((x+3)(x-2))
• x -intercept: 5
y = ((x-2)(x-5))/((x+3)(x-2))
• Horizontal asymptote: y=3/2
y = (3(x-2)(x-5))/(2(x+3)(x-2))
graph with your favorite utility to confirm.
To find the equation of a rational function with the given conditions, we can start by setting up the general form of a rational function:
f(x) = (ax + b)/(cx + d)
where a, b, c, and d are constants.
Given that the vertical asymptote is x = -3, we know that (cx + d) should be equal to zero at x = -3. Therefore, we have:
cx + d = 0 --> equation (1)
To find the value of c and d, we can substitute x = 5 into the rational function. Since the x-intercept is 5, we know that (ax + b) should be equal to zero at x = 5. Therefore, we have:
a(5) + b = 0 --> equation (2)
Now, we can determine the values of c and d by solving equations (1) and (2).
From equation (1), we have:
-3c + d = 0 --> equation (3)
From equation (2), we have:
5a + b = 0 --> equation (4)
Since we also know that the horizontal asymptote is y = 3/2, we can infer that the constant a/c should be equal to 3/2. Therefore, we have:
a/c = 3/2 --> equation (5)
Now, we can solve equations (3), (4), and (5) simultaneously to find the values of a, b, c, and d.
From equation (3), we have:
d = 3c --> equation (6)
Substituting equation (6) into equation (5), we have:
a/c = 3/2
a/(d/3) = 3/2
2a = 3d --> equation (7)
Now, substituting equation (7) into equation (4), we have:
5(3d/2) + b = 0
15d/2 + b = 0
b = -15d/2 --> equation (8)
Substituting equation (8) into equation (2), we have:
5a - 15d/2 = 0
10a - 15d = 0
2a - 3d = 0
2a = 3d
a = 3d/2 --> equation (9)
Finally, substituting equations (9) and (8) into the general form of the rational function, we have:
f(x) = (3d/2)x + (-15d/2) / (cx + d)
Now, to find the value of d, we can use the fact that the function has a hole at x = 2. This means that the denominator (cx + d) should be equal to zero at x = 2, but the numerator (3d/2)x + (-15d/2) should not be equal to zero at x = 2.
Therefore, we have:
2c + d = 0 --> equation (10)
(3d/2)(2) + (-15d/2) ≠ 0 --> equation (11)
Simplifying equation (11), we have:
3d + (-15d) ≠ 0
-12d ≠ 0
d ≠ 0
Now, substituting equation (10) into equation (11), we have:
(2c + (-2c)) ≠ 0
0 ≠ 0
Since 0 ≠ 0 does not hold true, there is no solution for d.
Therefore, we cannot find a single equation of a rational function that satisfies all the given conditions.
To find an equation of a rational function that satisfies the given conditions, we can start by considering the general form of a rational function:
f(x) = (a(x - h)(x - k)) / (b(x - p)(x - q))
Where:
- a, b are constants that affect the shape and stretching of the function.
- (h, k) represents the coordinates of any hole(s) in the graph of the function.
- (p, q) represents the coordinates of any x-intercept(s) in the graph of the function.
Given the conditions:
1. Vertical asymptote: x = -3
Since x = -3 is a vertical asymptote, it means that (x + 3) should be a factor in the denominator of the rational function.
2. Horizontal asymptote: y = 3/2
The horizontal asymptote y = 3/2 suggests that at the end, the highest degrees of the numerator and denominator should be equal to maintain the value of the fraction. Thus, the highest degree of the numerator and denominator should be the same.
3. x-intercept: 5
To have an x-intercept at x = 5, the factor (x - 5) should be a factor in the numerator.
4. Hole at x = 2
To have a hole at x = 2, (x - 2) should be a factor in both the numerator and the denominator.
Putting all these conditions together, we can form the equation:
f(x) = (a(x - h)(x - k)(x - 2)) / (b(x + 3)(x - p)(x - q))
Now, to determine the values of the constants a, b, h, k, p, q, we can use the given conditions. From the conditions, we have:
Vertical asymptote: x = -3
This suggests that (x + 3) is a factor in the denominator. So, p = -3.
x-intercept: 5
This suggests that (x - 5) is a factor in the numerator. So, k = 5.
Hole: x = 2
To have a hole at x = 2, (x - 2) should be a factor in both the numerator and the denominator. Since we already have (x - 2) in the numerator, it should also be a factor in the denominator. So, q = 2.
Horizontal asymptote: y = 3/2
For the horizontal asymptote, we look at the degree of the polynomials in the numerator and denominator. The degree of the numerator is one more than the degree of the denominator. So, to match the degrees and achieve the horizontal asymptote y = 3/2, we set a/b = 3/2. This gives us a = 3/2 * b.
Now, we can substitute these values into the equation:
f(x) = ((3/2 * b)(x - 5)(x - 2)(x - h)) / (b(x + 3)(x - p)(x - 2))
Simplifying further:
f(x) = (3/2)(x - 5)(x - 2)(x - h) / ((x + 3)(x + 2)(x + h - 2))
And there you have it, an equation of a rational function that satisfies the given conditions. Remember, the specific value of b and h have not been determined, so the equation has infinite possibilities.