Find a formula for the fourth degree polynomial p(x) whose graph is symmetric about the y-axis, and which has a y-intercept of 4, and global maxima at (1,6) and(-1,6).

Answer 1

y = a x^4 + b x^2 + c if symmetric about x = 0

when x = 0, y = 4
so
c = 4 and
y = a x^4 + b x^2 + 4

when x = 1, y = 6
6 = a + b + 4
so
a+b = 2
b = (2-a)

dy/dx = 4 a x^3 + 2 b x
= 0 when x = 1 or - 1

0 = 4 a + 2 b
0 = 4 a + 2(2-a)
0 = 4 a + 4 -2a
2 a = -4
a = - 2
b = 4

y = -2 x^4 + 4 x^2 + 4

Answer 2

the symmetry tells us that there is a local minimum at y=0

So, p'(x) = ax(x-1)(x+1) = a(x^3-x)
p(x) = a/4 x^4 - a/2 x^2 + c

since p(0) = 4,
p(x) = a/4 x^4 - a/2 x^2 + 4

Now, we know p(1)=p(-1) = 6, so
6 = a/4 - a/2 + 4
a = -8

so,
p(x) = -2x^4 + 4x^2 + 4