Thursday
March 23, 2017

Post a New Question

Posted by on Sunday, April 7, 2013 at 9:20pm.

Find a formula for the fourth degree polynomial p(x) whose graph is symmetric about the y-axis (meaning it has no odd powers of x), and which has a y-intercept of 10, and global maxima at (3,253) and (−3,253).

  • Calculus - , Sunday, April 7, 2013 at 10:46pm

    let the function be
    y = ax^4 + bx^2 + c
    since (0,10) is on it ----> c = 10

    dy/dx = 4ax^3 + 2bx
    at (3,253), dy/dx = 0
    4a(27) + 2b(3) = 0
    108a + 6b = 018a + b = 0
    b = -18a

    at (-3, 253) , dy/dx = 0
    4a(-27) + 2b(-3) = 0 , this gives the same result, so nothing new here

    but (3,253) lies on the actual curve
    253 = 81a + 9b
    sub in b = -18a
    253 = 81a + 9(-18a)
    253 = -81a
    a = -253/81
    then b = -18(-253/81) = 506/9

    so y = (-253/81)x^4 + (506/9)x^2 + 10


    check:
    dy/dx = -1012/81x^3 + 1012x = 0 for a max/min
    divide by -1012
    x^3/81 - x = 0
    times 81
    x^3 - 81x = 0
    x(x^2 - 9) = 0
    x = 0 , or x = ±3
    if x = 3 or -2
    y = (-253/81)(81) + (506/9)(9)
    = -253 + 506 =253 YEAHHH
    (weird coefficients)

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question