Question

How do you find a cubic function `y = ax^3+bx^2+cx+d` whose graph has horizontal tangents at the points(-2,6) and (2,0)?

Answer 1

`f(x)=3/16x^3-9/4 x+3`

Explanation:

Given `f(x)=ax^3+bx^2+cx+d` the condition of horizontal tangency at points `{x_1,y_1},{x_2,y_2}` is
`(df)/(dx)f(x=x_1) = 3ax_1^2+2bx_1+c=0`
`(df)/(dx)f(x=x_2) = 3ax_2^2+2bx_2+c=0`
also we have in horizontal tangency
`f(x=x_1)=ax_1^3+bx_1^2+cx_1+d = y_1`
`f(x=x_2)=ax_2^3+bx_2^2+cx_2+d = y_2`
so we have the equation system
# ((12 a - 4 b + c = 0), (12 a + 4 b + c = 0), (-8 a + 4 b - 2 c + d = 6), (8 a + 4 b + 2 c + d = 0)) #
Solving for `a,b,c,d` we get
`((a = 3/16), (b = 0), (c = -9/4), (d = 3))`

Answer 2

`y = 3/16x^3-9/4x+3`

Explanation:

Here's an alternative method without using differentiation.

Since there's a horizontal tangent at `(2, 0)`, there is a double zero there.

So: `f(x) = ax^3+bx^2+cx+d` is a multiple of:

`(x-2)(x-2) = x^2-4x+4`

Since the problem is symmetric, the cubic will also pass through the midpoint:

`((-2+2)/2, (6+0)/2) = (0, 3)`

and has rotational symmetry around it.

So `f(x) - 3` will be an odd function and have no `x^2` term.

`(ax+3/4)(x^2-4x+4) = ax^3+(3/4-4a)x^2+(4a-3)x+3`

So in order that there is no `x^2` term, we must have `a=3/16`

Hence:

`{ (a=3/16), (b=0), (c=4a-3 = 3/4-3 = -9/4), (d=3) :}`

`y = 3/16x^3-9/4x+3`

graph{3/16x^3-9/4x+3 [-10.21, 9.79, -1.44, 8.56]}