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)?

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)?

Emilia Aguilar · Accepted Answer

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

Scarlett Allison · Answer

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

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]}

Adam Adkins · Answer

undefined To find a cubic function with horizontal tangents at (-2,6) and (2,0), we need to determine the values of a, b, c, and d in the equation y = ax^3 + bx^2 + cx + d.

First, we know that the graph has horizontal tangents at (-2,6) and (2,0). This means that the derivative of the function at these points is equal to zero.

Taking the derivative of y with respect to x, we get:
y' = 3ax^2 + 2bx + c

Setting y' equal to zero at x = -2 and x = 2, we have two equations:
3a(-2)^2 + 2b(-2) + c = 0
3a(2)^2 + 2b(2) + c = 0

Simplifying these equations, we get:
12a - 4b + c = 0
12a + 4b + c = 0

Solving these equations simultaneously, we find:
a = 0, b = 0, c = 0

Therefore, the cubic function with horizontal tangents at (-2,6) and (2,0) is y = 0x^3 + 0x^2 + 0x + d, which simplifies to y = d. The value of d can be any real number.

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)?

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)?