Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 85

Find a cubic function $y = ax^3 + bx^2 + cx + d$ whose graph has horizontal tangent at the points $(-2,6)$ and $(2,0)$

If the graph has horizontal tangents at some point, it means that their slope is equal to 0 at that point.
And the first derivative of a function is equal to its slope.


$
\begin{equation}
\begin{aligned}
y &= ax^3 + bx^2 + cx +d \\
\\
y' &= a \frac{d}{dx} ( x^3) + b \frac{d}{dx} (x^2) + c \frac{d}{dx} (x) = \frac{d}{dx} (d)\\
\\
0 &= a(3x^2) + b(2x) + c(1) + 0\\
\\
0 &= 3x^2a+2xb+c\\
\phantom{x}& \text{when } (x=-2)\\
\\
0 &= 3(-2)^2 a + 2(-2)b+c\\
\\
0 &= 12a - 4b + c && \text{Equation 1}\\
\phantom{x} & \text{when}(x=2)\\
\\
0 &= 3 (2)^2 a + 2(2)b + c\\
\\
0 &= 12a + 4b + c && \text{Equation 2}
\end{aligned}
\end{equation}
$


We can get another 2 Equations by substituting the given points to the given solution
when $x = -2$, $y = 6$

$
\begin{equation}
\begin{aligned}
6 & = a ( -2)^3 + b(-2)^2 + c(-2) +d\\
6 & = -8a + 4b - 2c + d && \text{Equation 3}
\end{aligned}
\end{equation}
$


when $x= 2$, $y = 0$

$
\begin{equation}
\begin{aligned}
0 &= a(2)^3 + b(2)^2 + c(2) +d\\
0 & = 8a + 4b + 2c + d && \text{Equation 4}
\end{aligned}
\end{equation}
$

We have 4 Equations and 4 Unknowns, combining these equations we get.

$
\begin{equation}
\begin{aligned}
12a - 4b + c &= 0\\
\\
12a + 4b + c &= 0\\
\\
-8a+4b-2c+d &= 6\\
\\
8a+4b+2c+d &= 0
\end{aligned}
\end{equation}
$


$
\begin{equation}
\begin{aligned}
a &= \frac{3}{16}\\
b &= 0\\
c &= \frac{-9}{4}\\
d &= 3
\end{aligned}
\end{equation}
$


Therefore, the cubic equation is $\displaystyle y = \frac{3x^3}{16} - \frac{9}{4} x + 3$