# Find the cubic polynomial f(x) = ax^3 + bx^2 + cx + d that has horizontal tangents at the points (-4,5) and (3,2).

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to determine a polynomial f(x) = ax^3 + bx^2 + cx + d given that it has horizontal tangents at the points ( -4,5) and (3,2).

Now the slope of the tangent at a point on a curve can be derived from the first derivative.

The derivative of f(x) = ax^3 + bx^2 + cx + d is

f'(x) = 3ax^2 + 2bx + c.

At the points (-4, 5) and (3,2), the slope of the tangent is zero.

3*16*a - 2*4*b + c = 0

=> 48a - 8b + c = 0

3*9*a + 2*3*b + c = 0

=> 27a + 6b + c = 0

Also, as the points (-4,5) and (3,2) lie on the curve

5 = a*(-4)^3 + b(-4)^2 -4c + d

=> 5 = -64a + 16b - 4c + d

2 = a*3^3 + b*3^2 + c*3 + d

=> 2 = 27a + 9b + 3c + d

So we know have to solve these 4 equations:

27a + 9b + 3c + d = 2 ...(1)

-64a + 16b - 4c + d = 5 ...(2)

48a - 8b + c = 0 ...(3)

27a + 6b + c = 0 ...(4)

(1) - (2)

=> 91a - 7b + 7c = -3 ...(5)

a = 6/343 , b = 9/343 , c = -216/343 and d = 1091/343

Therefore the polynomial is f(x) = (6/343)x^3 + (9/343)x^2 - (216/343)x + 1091/343.

neela | High School Teacher | (Level 3) Valedictorian

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The tangents to y = f(x) is given by y-y1 = m(x-x1), where m = f'(x) at x= x1,

The given cubic is ax^3+bx^2+cx+d.

Therefore f'(x) = 3ax^2+2bx+c.

At (-4,5),  f'(x) = f'(-4) = 3a(-4)^2+2b(-4)+x = 48a-8b+c . But since the tangent is horizontal, f'(-4) = 0.

So 48a-4b+c = 0....(1).

The tangent at (3,2) is horizontal.

So f'(3) = 3a(2)^2+2b(2)+c = 0

So 12a+4b+C = 0...(2).

The point (-4,5) is on the cubic ax^3+bx^2+cx+d = y, so the point should satisfy the cubic equation:

-48a+16b-4x+d = 5...(3)

Also the point (3,2) is on ax^3+bx^2+cx+d = y. So we get:

27a+9b+3c+d = 2....(4).

If we solve the system of equations (1), (2), (3) and (4), we get the values of the 4 unknowns a, b, c and d.

(calculations not shown)

a = 6/343.

b = 9/343.

c = -216/343.

d = 1091/343.

Therefore y = ax^3+bx^2+cx+d = (3x^3+9x^2-216x+1091)/343.