Question

an equation of the line tangent to y = x^3 + 3x^2 + 2 at it's point of inflection is?

Answer 1

To find the equation of the tangent line to a curve at its point of inflection, we first need to find the x-coordinate of the point of inflection.

The point of inflection occurs where the second derivative of the function equals zero.

Given y = x^3 + 3x^2 + 2, we calculate the first derivative:

dy/dx = 3x^2 + 6x.

Now, we calculate the second derivative:

d^2y/dx^2 = 6x + 6.

To find the x-coordinate of the point of inflection, we set d^2y/dx^2 equal to zero and solve for x:

6x + 6 = 0
6x = -6
x = -1.

The x-coordinate of the point of inflection is -1.

To find the y-coordinate of the point of inflection, we substitute x = -1 into the original equation:

y = (-1)^3 + 3(-1)^2 + 2
y = -1 + 3 + 2
y = 4.

The y-coordinate of the point of inflection is 4.

So, the point of inflection is (-1, 4).

To find the equation of the tangent line at this point, we need the slope. The slope of the tangent line is equal to the first derivative evaluated at the point (-1, 4).

dy/dx = 3(-1)^2 + 6(-1)
dy/dx = 3 - 6
dy/dx = -3.

The slope of the tangent line is -3.

Using the point-slope form of a linear equation, we can write the equation of the tangent line:

y - y1 = m(x - x1),

where (x1, y1) is the point of inflection and m is the slope.

Substituting the known values, we have:

y - 4 = -3(x - (-1))
y - 4 = -3(x + 1)
y - 4 = -3x - 3
y = -3x + 1.

Therefore, the equation of the line tangent to y = x^3 + 3x^2 + 2 at its point of inflection is y = -3x + 1.