Question

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

Answer 1

Answer:

$y = -3x +1$

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:

$y" = 6x + 6$

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: $y = -3x +1$

Answer 2

Answer:

y=-3x+1

Explanation: