create a p(x) with the following conditions:
A) Vertical asymptotes at x=5 and x=-2
B) A double root x=-3
C) A horizontal asymptote at y=3
(A) y = 1/(x-2)(x-5)
(B) y = (x+3)^2 / (x-2)(x-5)
(C) y = 3(x+3)^2 / (x-2)(x-5)
(3x+9)(x+3)/ [(x-5)(x+2)]
in the numerator I used 3x+9 instead of x+3 so that as x goes to infinity
the fraction goes to 3x^2 + little stuff /x^2 + little stuff ---> 3
Sorry - I missed the -2 ; read x=2
To create a polynomial P(x) that satisfies the given conditions, we can use the following step-by-step process:
A) Vertical asymptotes at x=5 and x=-2:
Vertical asymptotes occur when the denominator of a rational function equals zero. Since we want vertical asymptotes at x=5 and x=-2, we can include the factors (x - 5) and (x + 2) in the denominator. Therefore, the denominator of our function will be: (x - 5)(x + 2).
B) A double root at x=-3:
A double root at x=-3 means that (x + 3) will be a factor raised to the power of 2 in our polynomial function. Therefore, we can include (x + 3)^2 in our polynomial.
C) A horizontal asymptote at y=3:
To achieve a horizontal asymptote at y=3, the leading term of our polynomial should be a constant multiple of x^0 (which is 1). Therefore, the numerator of our polynomial P(x) can be 3.
Considering all the conditions, we can formulate our polynomial as:
P(x) = 3(x + 3)^2 / ((x - 5)(x + 2))
This polynomial satisfies the given conditions:
A) It has vertical asymptotes at x=5 (due to (x - 5)) and x=-2 (due to (x + 2)).
B) It has a double root at x=-3 (due to (x + 3)^2).
C) It has a horizontal asymptote at y=3 (as the degree of the denominator is greater than the numerator).
Please note that this is just one possible solution, and you can create different polynomials that satisfy the given conditions.