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maxonik [38]
3 years ago
10

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

Advanced Placement (AP)
2 answers:
seraphim [82]3 years ago
4 0

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

<em>Hence, the equation is: </em>y = -3x +1<em></em>

daser333 [38]3 years ago
3 0

Answer:

y=-3x+1

Explanation:

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