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March 5, 2015

March 5, 2015

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

- Calculus -
**Reiny**, Sunday, April 7, 2013 at 10:46pmlet 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)

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