Find the cubic function whose graph has horizontal Tangents Problem: Find the cubic function $y = ax^3 + bx^2 + cx + d$ whose graph has horizontal tangents at $(-2, 6)$ and $(2, 0)$.
Now I can never seem to gather enough information for find all the values of $a,b,c,d$.
All the information I can gather is:


*

*the derivative $\dfrac{\mathrm{d}y}{\mathrm{d}x}$.

*$\dfrac{\mathrm{d}y}{\mathrm{d}x}$ is zero at those points, and therefore can set up two equations for the derivative.

*Those given points lie on the graph and therefore I can plug in the $x$ and $y$ values of both points into the original equation set up two more equations.
Even with all this information I was only able to find out the $b = 0$, and just don't know what to do from there!
 A: From the fact that $\frac{dy}{dx}=0$ at $x=-2$ and $x=2$, we get
$$f'(x)=k(x+2)(x-2)=kx^2-4k$$
for a scaling constant $k$.
Then by taking the anti-derivative
$$f(x)=\frac{1}{3}kx^3-4kx+C$$
Plugging in the coordinates of the two points on the curve
$$\begin{array}{ccccc}
f(-2)&=&-\frac{8}{3}k+8k+C&=\frac{16}{3}k+C=6 \\
f(2)&=&\frac{8}{3}k-8k+C&=-\frac{16}{3}k+C=0
\end{array}$$
with solution $k=\frac{9}{16},C=3$.
So $\boxed{f(x)=\frac{3}{16}x^3-\frac{9}{4}x+3}$
A: Here’s yet another approach: the graph of a cubic polynomial $f$ has rotational symmetry (180°) about the inflection point, so if we place the local max and min symmetrically about the origin, we get an odd cubic polynomial. You make this move by subtracting $3$ from the function, and I’ll call the result $g(x)=f(x)-3$.
We now have $g(x)=ax^3+bx$, $g(2)=-3$, $g'(2)=0$, giving the linear equations $8a+2b=-3$, $12a+b=0$, $b=-12a$, $8a-24a=-3$, $a=3/16$, $b=-9/4$. Thus we have $g(x)=3x^3/16-9x/4$, $f(x)=3x^3/16-9x/4+3$.
A: Hint:
$$y'=3ax^2+2bx+c$$
at (-2,6) the slope equal to the slope at (2,0)
$$3a(-2)^2+2b(-2)+c=3a(2)^2+2b(2)+c$$
$$b=0$$
to complete the solution, we will use the $y'=0$ at point (-2,6) or (2,0)(if you use two point, you will get same equation)
$$3a(-2)^2+c=0$$
now use the two points with the original equations
$$a(-2)^3+c(-2)+d=6$$
$$a(2)^3+c(2)+d=0$$
solve the three equations to get the coefficients
A: You have the informations:
$$
y'(-2)=0\quad y'(2)=0 \quad y(-2)=6 \quad y(2)=0
$$
that is: four equations for the four unknowns $a,b,c,d$
$$
\begin{cases}
12a-4b+c=0\\
12a+4b+c=0\\
-8a+4b-2c+d=6\\
8a+4b+2c+d=0
\end{cases}
$$
Can  you solve this system ?

Find : $a=3/16$, $b=0$, $c=-9/4$, $d=3$.

A: Hint:
With $y'(-2)$ = 0 and $y'(2) = 0$ you get $b = 0$ and $12a+c = 0$
With $y(2) = 6$ and $y(-2) = 0 $ you get $d = 3$ and $8a + 2c = -3$
Now take the equations of the above two and get a and c
A: The derivative must be a multiple of $x^2-4$, hence the polynomial is $A\left(\dfrac{x^3}3-4x\right)+B$.
Then $P(2)+P(-2)=2B=6$, and $P(2)-P(-2)=2A\left(\dfrac83-8\right)=-6$, and
$$P(x)=\frac9{16}\left(\frac{x^3}3-4x\right)+3.$$
