Question

An equation of the line tangent to y=x3+3x2+2 at its point of inflection is

A. y=−6x−6y is equal to negative 6 x minus 6

B. y=−3x+1y is equal to negative 3 x plus 1

C. y=2x+10y is equal to 2 x plus 10

D. y=3x−1

Answer 1

`y = -3x +1`

Step-by-step explanation:

Given

`y = x^3 + 3x^2 + 2`

Required

Determine the equation at the point of inflection

The point of inflection of a curve is the point where the second derivative is 0.

So:

`y = x^3 + 3x^2 + 2`

First derivative is:

`y' = 3x^2 + 6x`

Second derivative:

`y`

Equate to 0

`6x + 6 = 0`

Solve for x

`6x = -6`

`x = -1`

Next, we calculate the slope (m) of the point using the first derivative.

Substitute -1 for x in
`y' = 3x^2 + 6x`

`m = 3*-1^2 + 6 * -1`

`m = 3*1 - 6`

`m = 3 - 6`

`m = -3`

To get the y equivalent;

Substitute
`x = -1` in
`y = x^3 + 3x^2 + 2`

`y = (-1)^3 + 3(-1)^2 + 2`

`y = -1 + 3+ 2`

`y = 4`

Lastly, we calculate the line equation using:

`y - y_1 =m(x- x_1)`

Where

`x = -1` and
`y = 4`

`m = -3`

So, we have:

`y - 4 = -3(x -(-1))`

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

`y - 4 = -3x -3`

Make y the subject

`y = -3x -3 +4`

`y = -3x +1`

Hence, the equation is:
`y = -3x +1`