calclus question....tell me in detail?

Question

3.Determine the equation of and graph a polynomial function of degree 4 with the following properties:
a)4 x intercepts
b)2 local max values and 1 local min
c)Goes through either (2,1), (4,6), or (-1,-3)
d)Has at least 2 points of inflection
e)Has no asymptotes

Be sure to show appropriate calculations and justifications for all of your steps. First and second derivative tests will be required on your final function to justify your choices.

Answer 1

Polynomials never have asymptotes, so e is a freebie. If it has four x-intercepts then it automatically has 3 critical points (Rolle's thm). To make sure two are maxes and one is a min, just make the leading coefficient negative. 3 local extrema gaurantee two inflection points. So that takes care of a&b&d&e. Since (c) is "or" but not "and" I can pick and choose. ("AND" might be a little more difficult.) Easiest way is to make sure that the point (-1,-3) is part of the graph that is on its way down -> -infinity, i.e., after all the roots. So set up f(x)=k (x+5)(x+4) (x+3)(x+2) and then choose k so that f(-1)=-3 so k=-1/8 and that should do it. (-1/8) (x+5) (x+4) (x+3) (x+2)

Critical pts at x=-7/2, -7/2+1/2*5^(1/2), -7/2-1/2*5^(1/2) and inflection points at x= -7/2+1/6*15^(1/2), -7/2-1/6*15^(1/2)

Answer 2

Let f(x) = -1/40(x^2 - 9)(x^2 - 16) = -1/40(x-3)(x+3)(x-4)(x+4)

a) The 4 x-intercepts occur when f(x) = 0, at x = -3,+3, -4, +4

Differentiating

f '(x) = -1/20(x)(2x^2 - 25)

f"(x) = -1/20(6x^2 - 25)

Solving

f '(x) = 0 at x = 0, 5/sqrt(2), -5/sqrt(2)

f"(x) = 0 at x = 5/sqrt(6), -5/sqrt(6).

f(x) is concave down for x < -5/sqrt(6) and
x > 5/sqrt(6) and concave up for the interval
-5/sqrt(6) < x < 5/sqrt(6).

b) Therefore local maxima occur at x = 5/sqrt(2) and
x = -5/sqrt(2) since the second derivative is negative at these critical points, and a local minimum occurs at x = 0 since the second derivative is positive at the critical point x = 0.

c) f(-1) = -1/40(-8)(-15) = -3, so the graph goes through (-1,-3).

d) The points of inflection are at x = -5/sqrt(6) and x = 5/sqrt(6).

The graph is symmetric about zero, so it should be easy to draw from the above information.

e) Since x gets arbitrarly large and negative as x gets large in either direction, there are no asymptotes.