CALCULUS HELP PLEASE!!!!?

Question

What is the equation of a line tangent to the graph of y= x^(3) + 3x^(2) + 2 at its point of inflection?

Please help I'm having so much trouble

Answer 1

y = x³ + 3x² + 2

The point of inflection occurs when y'' = 0.

Apply derivative with respect to x each side.

(d/dx)y = (d/dx)(x³ + 3x² + 2)

y' = (d/dx)(x³) + (d/dx)(3x²) +(d/dx)( 2)

Recall Power Rule for Function is (d/dx)(xⁿ) = n x^n-1

General Derivative Formula is (dc/dx) = 0 where c is constant

y' = 3x² + 3(2x) + 0

y' = 3x² + 6x

Again apply derivative with respect to x each side.

y'' = 6x + 6

If y'' = 0 then 6x + 6 = 0

So x + 1 = 0

Subtract 1 from each side.

x = -1

When x = -1, y = (-1)³ + 3(-1)² + 2 = -1 + 3 + 2 = 4

 So the inflection point is (-1, 4).

 

At x = -1, the slope of the tangent line is y' = 3(-1)² + 6(-1) = 3 - 6 = -3.

Recall point slope form is y - y₁ = m (x - x₁)

Point(x₁, y₁) = (-1, 4) and slope m = - 3

So the equation of the tangent line at the point of inflection is:

y - 4 = -3(x + 1)

y - 4 = - 3x - 3

Add 4 to each side

y = - 3x + 1

The equation of the tangent line at the point of inflection is y = - 3x + 1