If it is even, P(x)=ax^4+bx^2+c
P(0)=10 so c is 10
p'=5ax^3+2bx=0
x(5ax^2+2b)=0
maximum at x=+-1
so 5ax^2=-2b
or 5a=-2b or a=-5/2 b
but P(1)=13=(-5/2 b)*1+ b(1)+10
or 3/2 b=10 b=20/3 and a=-5/2 * 20/3=100/6
check P(-1)=13
it checks.
fourth degree polynomial p(x) whose graph is symmetric about the y-axis, and which has a y-intercept of 10, and global maxima at (1,13) and (-1, 13).
3 answers
Why do we end up with f(x) instead of f(t) on the Second Fundamental Theorem of Calculus?
5x^2