Question 979388
f(4)=6

If there is a zero at 4+2 sqrt (3), there is a zero at 4 - 2 sqrt (3)
a*4+b*2+c=4
4a+2b+c=4

16a+4b+c=6
Eliminate c
4a+2b+c=4
-16a-4b-c=-6
-12a-2b=-2
12a+2b=2
6a+b=1
But -b/2a=4, so -b=8a
-2a=1
a=(-1/2)
b=4
first equation -2+8+c=4; c=-2
second equation -8+16+c=6; c=-2

(-1/2)x^2+4x-2=f(x)
quadratic formula:
{-4 +/- sqrt (16-4)}/-1
roots are 4+/- sqrt (12); sqrt (12=2 sqrt (3))

a= -1/2
b=4
c= -2

{{{graph(300,300,-10,10,-10,10,(-1/2)x^2+4x-2)}}}