[Math] Find the cubic function whose graph has horizontal Tangents

calculusderivativessystems of equations

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:

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

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

  3. 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!

Best Answer

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 $\boxed{f(x)=\frac{3}{16}x^3-\frac{9}{4}x+3}$

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