Posted by **Cad** on Sunday, April 3, 2011 at 4:15pm.

Find the equation of a quartic polynomial whose graph is symmetric about the y-axis and has local maxima at (−1,1) and (1,1) and a y-intercept of -1.

- Calculus -
**Damon**, Sunday, April 3, 2011 at 5:03pm
y = a x^4 + b x^3 + c x^2 + d x + e

when x = 0, y = -1 so e = -1

so

y = a x^4 + b x^3 + c x^2 + d x - 1

for max or min

dy/dx = 0 = 4 a x^3 + 3 b x^2 + +2 c x + d

when x = -1

0 = 4 a (-1) +3b(1)+2c(-1)+d

1 = a (1) + b(-1) + c(1) +d(-1) - 1

when x = +1

0 = 4a + 3 b + 2c + d

1 = a+b+c+d-1

--------------------------------

start solving

0 = -4a+3b-2c+d

0 = +4a+3b+2c+d

-----------------

0 = 6b +2d

2 = a-b+c-d

2 = a+b+c+d

-------------

0 = -2b-2d

0 = 6b +2d

0 = -2b-2d

-----------

8 b = 0

b=0

then d = 0

(in fact from symmetry I bet all odd terms are 0)

now go back and get a and c

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