How to Derive a Cubic Function with Horizontal Tangents at Given Points?

[karush]

$$y=ax^3+bx^2+cx+d$$

Find the cubic eq whose graph has horz tangent at the points$ (-2,6)$ and $ (2,0)$
$$y'=3a{x}^{2}+2bx+c$$

 

[evinda]

$$y=ax^3+bx^2+cx+d$$

Find the cubic eq whose graph has horz tangent at the points$ (-2,6)$ and $ (2,0)$
$$y'=3a{x}^{2}+2bx+c$$

(Wave)

We have $y=f(x)=ax^3 + bx^2 + cx + d $ .

We want to find the values of $a, b, c$ and $d$.

We know that $f(-2)=6$, so we have: $-8a + 4b - 2c + d = 6 $.

We also know that $f(2)=0$, so we have: $8a+4b+2c+d=0$.

Since there are horizontal tangents at those two points, we know that the derivative is zero at those points, i.e. $f'(-2)=f'(2)=0$ .

Then you will have 4 equations and you will solve the system.

 

[karush]

Isn't $d=3$

Of which we only need a, b, c

 

[evinda]

Isn't $d=3$

Of which we only need a, b, c

Yes, it is $d=3$.

Making operations, we can find for example the equalities $16a+4c=-6$ and $12a+c=0$ and we deduce that $a=\frac{3}{16}$.

 

[karush]

Then $c=-\frac{9}{4}$

So we have b left

So then $$f'(-2)=f'(2)=0$$

Is next?

 

[evinda]

Then $c=-\frac{9}{4}$

Exactly! (Nod)

So we have b left

So then $$f'(-2)=f'(2)=0$$

Is next?

You said previously that $d=3$. Then by adding the equality that we get from $f(2)=6$ with the equality we get from $f(-2)=0$, if we multiply it by $-1$ , we have that $4b+d=3$. So... ?

 

[karush]

$$y=\frac{3}{16}{x}^{3}-\frac{9}{4}x+3$$

I hope

 

[evinda]

$y=\frac{3}{16}{x}^{3}-\frac{9}{4}x+3$$

I hope

Well done! (Clapping)