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.
Well first lets get one with three real zeros.
y = k(x-x1)(x-x2)(x-x3)
but one happens twice
y = k(x-x1)(x-x2)(x-x2)
denominator = 0 at x=-2 and x = 3
y = c[(x-x1)(x-x2)(x-x2)]/[(x+2)(x-3)]
when x = 0, y = 1
y =1 = c[-x1 x2^2]/-6
6 = c x1 x2^2
when x and y are big, y = 2x
big x
y = c x^3/x^2 = c x
so c = 2
so
x1 x2^2 = 3 or x1 = x2^2/3
y = 2[(x-x1)(x-x2)(x-x2)]/[(x+2)(x-3)]
y = 2[(x-x2^2/3)(x-x2)(x-x2)]/[(x+2)(x-3)]
hmmm, unless I am missing something I am still free to choose x2
x1 x2^2 = 3 or x1 = 3/x2^2
y = 2[(x-x1)(x-x2)(x-x2)]/[(x+2)(x-3)]
y = 2[(x - 3/x2^2)(x-x2)(x-x2)]/[(x+2)(x-3)]
To design a rational function with the given characteristics, we need to consider the factors that determine its equation.
First, let's consider the zeros of the function. We know that we have three real zeros, one of multiplicity 2. Therefore, our rational function will have the following factors: (x - a), (x - a), and (x - b), where a and b are constants representing the zeros.
Next, let's consider the vertical asymptotes. We know that the rational function has vertical asymptotes at x = -2 and x = 3. To achieve this, we need the factors (x + 2) and (x - 3) in the denominator.
Now, let's consider the oblique asymptote. We know that the oblique asymptote is given by the equation y = 2x + 1. To incorporate this, we want the degree of the numerator to be one more than the degree of the denominator.
Putting all this together, the rational function can be expressed as:
f(x) = (x - a)(x - a)(x - b) / ((x + 2)(x - 3)) + (2x + 1)
Since the y-intercept is 1, we can set f(0) = 1:
1 = (0 - a)(0 - a)(0 - b) / ((0 + 2)(0 - 3)) + (2(0) + 1)
Simplifying this equation will allow us to determine the values of a and b:
1 = a^2(b) / (-6) + 1
0 = a^2(b) - 6
At this point, we can choose arbitrary values for a and b such that the equation satisfies the given conditions. For example, let's set a = -1 and b = 6. Substituting these values back into our rational function, we get:
f(x) = (x + 1)(x + 1)(x - 6) / ((x + 2)(x - 3)) + (2x + 1)
Therefore, a possible rational function with the given characteristics is:
f(x) = (x + 1)(x + 1)(x - 6) / ((x + 2)(x - 3)) + (2x + 1)
To design a rational function with the given characteristics, we can start by writing the function in factored form using the zeros and their multiplicities. Let's suppose the three real zeros are a, b, and c, where b has a multiplicity of 2. So, the factored form of the function is:
f(x) = k * (x-a) * (x-b)^2 * (x-c)
Now, let's use the given information about the y-intercept, vertical asymptotes, and oblique asymptote to determine the values of a, b, c, and k.
1. Y-intercept: The y-intercept tells us that when x=0, f(x)=1. Plugging these values into the equation:
f(0) = k * (0-a) * (0-b)^2 * (0-c) = 1
Simplifying, we get:
k * (-a) * (b^2) * (-c) = 1
abc * k = -1 (Eq. 1)
2. Vertical asymptotes: We know that the vertical asymptotes are at x=-2 and x=3. This means that x=-2 and x=3 are not zeros of the function. Therefore, we can write two equations based on this information:
(x + 2) must be in the denominator of f(x), so (x + 2) is not a factor.
(x - 3) must be in the denominator of f(x), so (x - 3) is not a factor.
This implies that (x + 2) and (x - 3) are factors in the denominator. So, our factored form now looks like:
f(x) = k * (x - a) * (x - b)^2 * (x - c) / ((x + 2) * (x - 3))
3. Oblique asymptote: The oblique asymptote is given by the function y=2x+1. This means that when x approaches positive or negative infinity, the function f(x) approaches 2x+1. So, as x approaches infinity:
f(x) / x = (2x + 1) / x
To match this behavior, we need the leading terms in the numerator and denominator to be the same. The leading terms in the numerator are (x - a) * (x - b)^2 * (x - c) and the leading term in the denominator is (x + 2) * (x - 3). Therefore, we can write:
lim(x->infinity) [(x - a) * (x - b)^2 * (x - c)] / [(x + 2) * (x - 3)] = 2
Expanding the numerator and simplifying, we get:
lim(x->infinity) [(x^4 - 2(b+1)x^3 + (2b+b^2-2a-3)x^2 + (-2b^2a-ab^2-4ab+3b^2+3a^2+6)x + (a+b^2a-3ba^2))] / [(x^2 - x - 6)] = 2
To match the leading terms, we can ignore the lower-degree terms on both sides. So, we have:
lim(x->infinity) (x^4) / (x^2) = 2
Therefore, the leading coefficient of the numerator and denominator must be the same, which gives us:
1 = 2
This is not possible. So, we need to adjust our previous assumption.
Since the desired oblique asymptote is y=2x+1, we can modify the original assumption to:
f(x) = (2x + 1) * k * (x-a) * (x-b)^2 * (x-c) / [(x + 2) * (x - 3)]
Now, let's revisit the y-intercept (Eq. 1) and solve for k:
abc * k = -1
With a, b, and c being the zeros, let's say a = 0, b = p, and c = q
0 * p^2 * q * k = -1
pq * k = -1
As we established earlier, the y-intercept is (0, 1). Plugging this into the function:
f(0) = 1
(2(0) + 1) * k * (0 - 0) * (0-p)^2 * (0-q) / [(0 + 2) * (0 - 3)] = 1
k * p^2 * q / 6 = 1
pq * k = 6
Now we have a system of equations:
pq * k = -1
pq * k = 6
Solving this system, we find that k = -1/6, p = -2, and q = 3.
Finally, the rational function with the given characteristics is:
f(x) = (-2x + 1) * (-1/6) * (x - 0) * (x - (-2))^2 * (x - 3) / [(x + 2) * (x - 3)]