Design a rational function with the following characteristics: three real zeros, one of multiplicity 2; y-intercept 1; vertical asymptotes x=-2 and x=3; oblique asymptote y=2x+1.
First, we note that the oblique asymptote of a rational function has the form $y=mx+b$ where $m$ is the coefficient of the highest degree term of the polynomial in the numerator divided by the coefficient of the highest degree term of the polynomial in the denominator. Since the oblique asymptote is $y=2x+1$, this means that the numerator has a degree of 1 higher than the denominator.
Next, we note that the function will have vertical asymptotes at $x=-2$ and $x=3$ since these values cannot be in the domain of the function. This means that the denominator must have factors of $(x+2)$ and $(x-3)$.
Finally, since the function has a y-intercept of 1 and a zero of multiplicity 2 at some value $a$, the numerator must have a factor of $(x-a)^2$.
We can then write the function as follows:
[f(x) = \frac{(x-a)^2}{(x+2)(x-3)}]
To find the value of $a$, we can plug in the y-intercept of 1 and solve for $a$:
[1 = \frac{(a-a)^2}{(1+2)(1-3)} = \frac{a^2}{-4}]
Solving for $a$, we get $a = \pm 2i$.
Thus, the function that satisfies all the given conditions is:
[f(x) = \frac{(x-2i)^2}{(x+2)(x-3)}]
Why did the rational function go to the circus? Because it wanted to visit the "zero" of the fun!
Let's call our rational function f(x). Since we know that it has three real zeros, one of multiplicity 2, we can express it as:
f(x) = k(x - a)(x - b)^2(x - c)
Here, a, b, and c represent the three real zeros. We need the multiplicity of one zero to be 2, so we use (x - b)^2.
Since the y-intercept is 1, we know that f(0) = 1. Plugging in x = 0 into our function:
1 = k(0 - a)(0 - b)^2(0 - c)
Simplifying this, we can set a = 0, b = ±1, and c = 0 so that the y-intercept is 1.
Next, we know that the vertical asymptotes are x = -2 and x = 3. This means that (x - (-2)) and (x - 3) appear as factors in the denominator:
f(x) = k(x)(x + 2)(x - 3)^2
Finally, we have an oblique asymptote, given by y = 2x + 1. This means that, as x approaches ±infinity, the function will behave like the equation of the oblique asymptote. Let's focus on the behavior for x → ±infinity.
Expanding the function, we get:
f(x) = k(x)(x^3 - 7x^2 + 15x - 18)
As x approaches ±infinity, the higher-order terms become much larger than the lower-order terms. Thus, we can ignore the terms multiplied by x in comparison to x^3:
As x approaches ±infinity, f(x) ≈ k(x^3 - 7x^2 + 15x - 18)
Comparing this to the oblique asymptote (y = 2x + 1), we need the coefficients of the corresponding powers of x to match. For instance, the coefficient of x^3 in f(x) must be 2.
Therefore, 2 = k, and we have:
f(x) = 2x(x + 2)(x - 3)^2
Voila! Our circus-inspired rational function, with three real zeros, y-intercept 1, vertical asymptotes at x = -2 and x = 3, and an oblique asymptote y = 2x + 1.
To design a rational function with the specified characteristics, we'll start by considering its zeros and multiplicities. Let's say the three real zeros are a, b, and c, where b has a multiplicity of 2.
The fact that a function has zeros at a, b, and c means that it can be written in factored form as:
f(x) = A(x-a)(x-b)(x-c)
Since b has a multiplicity of 2, we have:
f(x) = A(x-a)(x-b)^2(x-c)
To find the value of A, we can use the y-intercept, which is given to be 1. This means that when x = 0, y = 1. Substituting these values into the equation, we get:
1 = A(0-a)(0-b)^2(0-c)
Simplifying, we have:
1 = A(-a)(-b)^2(-c)
1 = Aabc
So, A = 1/(abc).
Now, let's work on the vertical asymptotes. The vertical asymptotes occur where the denominator of the rational function is zero. We're given that the vertical asymptotes are x = -2 and x = 3. Therefore, the denominator should contain the factors (x+2) and (x-3). Adding these factors to our equation, we have:
f(x) = A(x-a)(x-b)^2(x-c)(x+2)(x-3)
For the oblique asymptote, we're given that it is y = 2x + 1. This means that as x approaches positive or negative infinity, the function approaches the line y = 2x + 1. To achieve this, the degree of the numerator should be one greater than the degree of the denominator.
To make the degree of the numerator one greater, we can multiply the entire equation by a linear factor in the form of (mx + n). This will give us:
f(x) = A(mx + n)(x-a)(x-b)^2(x-c)(x+2)(x-3)
Expanding this out, we have:
f(x) = A(mx^2 + nx - max - na)(x-b)(x-b)(x-c)(x+2)(x-3)
Simplifying further, we have:
f(x) = A(mx^3 + (n-ma)x^2 + (-na-nb+2ma)x + (-2ma-3na))
Now, we can equate the coefficients of this polynomial to the coefficients of the oblique asymptote equation y = 2x + 1 to find the values of m and n.
Comparing coefficients, we get:
A = 2m
n - ma = 1
-na - nb + 2ma = 2
-2ma -3na = 1
Solving these equations together will give us the values of m and n.
Once we have the values of A, a, b, c, m, and n, we can substitute them back into the equation:
f(x) = A(mx^3 + (n-ma)x^2 + (-na-nb+2ma)x + (-2ma-3na))
This equation represents a rational function with three real zeros, one of which has a multiplicity of 2. It has a vertical asymptote at x = -2 and x = 3, and an oblique asymptote at y = 2x + 1.
To design a rational function with the given characteristics, we need to consider the properties of rational functions and use the given information to construct the equation.
1. Real Zeros:
We are given that the function has three real zeros, one of which has multiplicity 2. Let's denote these zeros as a, b, and c. The equation for the rational function can be written in the factored form as:
f(x) = A(x - a)(x - b)(x - c)
2. Y-Intercept:
The function has a y-intercept of 1, which means it passes through the point (0, 1). Plugging in the values, we have:
f(0) = A(0 - a)(0 - b)(0 - c) = 1
This equation will help us determine the value of the constant A.
3. Vertical Asymptotes:
The function has vertical asymptotes at x = -2 and x = 3, which means the denominator of the rational function is zero at these points. We can include these conditions by including the factors (x + 2) and (x - 3) in the denominator:
f(x) = A(x - a)(x - b)(x - c)/[(x + 2)(x - 3)]
4. Oblique Asymptote:
The function has an oblique asymptote y = 2x + 1. This means that as x approaches positive or negative infinity, the function approaches the line with a slope of 2 and a y-intercept of 1.
To have this oblique asymptote, we divide the numerator by the denominator and take the limit as x approaches infinity:
lim(x -> ∞) [A(x - a)(x - b)(x - c)] / [(x + 2)(x - 3)] = 2x + 1
Simplifying and dividing numerator by x^3, we get:
lim(x -> ∞) [A(x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc)] / (x^3 + ...x^2...) = 2x + 1
Comparing the coefficients of x in the numerator and denominator, we have:
A = 2
a + b + c = 0 (since the coefficient of x^2 in the numerator is 0)
ab + ac + bc = 0 (since the coefficient of x in the numerator is 0)
abc = 0 (since the constant term in the numerator is 0)
These equations help us determine the values of a, b, and c.
Finally, substituting the value of A (which we found to be 2) and the values of a, b, and c into the equation, we get the rational function that satisfies all the given characteristics:
f(x) = 2(x - a)(x - b)(x - c)/[(x + 2)(x - 3)]