Answer
\[y=\frac{3}{16}x^3-\frac{9}{4}x+3\]
Work Step by Step
$y=ax^3+bx^2+cx+d$
$y$ has horizontal tangent if $y'(k)=0$
$y'=3ax^2+2bx+c $ ______(1)
According to question, $y$ has horizontal tangents at $(-2,6)$ and $(2,0)$
\[\Rightarrow y(-2)=6\;,\; y(2)=0\:,\:y'(-2)=0\:,\:y'(2)=0\]
By (1)
$y'(-2)=3a(-2)^2+2b(-2)+c$
$y'(-2)=12a-4b+c=0$ ____(2)
$y'(2)=3a(2)^2+2b(2)+c$
$y'(2)=12a+4b+c=0$ ____(3)
(3)$-$(2)
$\Rightarrow 8b=0\Rightarrow \underline{b=0}$
From (2)
$12a+c=0$ ____(4)
$y(-2)=a(-2)^3+b(-2)^2+c(-2)+d=6$
$\Rightarrow y(-2)=-8a+4b-2c+d=6$
$\Rightarrow -8a-2c+d=6$ ____(5)
$y(2)=a(2)^3+b(2)^2+c(2)+d=0$
$\Rightarrow y(2)=8a+4b+2c+d=0$
$\Rightarrow 8a+2c+d=0$ ____(6)
(5)+(6)
$\Rightarrow 2d=6\Rightarrow \underline{d=3}$
From (6)
$8a+2c=-3$ ____(7)
Multiply (4) by -2
$-24a-2c=0$ ___(8)
(7)+(8)
$-16a=-3\Rightarrow \underline{a=\frac{3}{16}}$
From (4)
$c=-12\left(\frac{3}{16}\right)=\underline{\frac{-9}{4}}$
Hence using values of $a,b,c $ and $d$
$y=\frac{3}{16}x^3-\frac{9}{4}x+3$.
