Answer
$f(x)=0.1875x^3-2.25x+3$
Work Step by Step
A function $f(x)$ has a horizontal function at $x=c$ if $f'(c)=0$
$f'(x)=3ax^{2}+bx+c$
Since we have to find $a, b, c,$ and $d$, we need 4 equations.
Setting $f'(x)=0$
At point $(-2,6)$ : $f'(-2)=3a(-2)^{2}+b(-2)+c = 0$
So our first equation is $12a-2b+c=0$
At point $(2,0)$ : $f'(2)=3a(2)^{2}+b(2)+c=0$
So our second equation is $12a+2b+c=0$
Setting $f(x)=y$
At point $(-2,6)$ : $f(-2)=a(-2)^3+b(-2)^2+c(-2)+d=6$
So our third equation is $-8a+4b-2c+d=6$
At point $(2, 0)$ : $f(2)=a(2)^3+b(2)^2+c(2)+d=0$
So our fourth equation is $8a+4b+2c+d=0$
So our equations are:
1) $12a-2b+c=0$
2) $12a+2b+c=0$
3) $-8a+4b-2c+d=6$
4) $8a+4b+2c+d=0$
Subtracting 1) - 2) we get the following:
$-2b-2b=0$ or $-4b=0$, by dividing both sides by -4, we conclude that $b=0$
Adding 3) + 4) we get the following:
$8b+2d=6$, since $b=0$, we get $2d=6$, dividing both sides by $2$, we can conclude that $d=3$
If we solve Eq. 1) for $a$, we get $a=-\frac{c}{12}$
Substituting $a=-\frac{c}{12}$ in Eq. 3) we get the following:
$-8(-\frac{c}{12})+2c+3=0$, by solving for $c$ we get that $c=-\frac{9}{4}$ or $-2.25$
If we substitute all the variables we obtained in any of the four equations, we conclude that $a=3/16$ or $0.1875$
So our function is:
$f(x)=0.1875x^3-2.25x+3$
