an equation of the line tangent to y = x^3 + 3x^2 + 2 at it's point of inflection is?
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.
To find the equation of the line tangent to the given function at its point of inflection, we need to find the derivative of the function and set it equal to zero to find the x-coordinate of the point of inflection. Then, we can substitute this value into the original function to find the corresponding y-coordinate. Finally, we can use the point-slope form of a line to determine the equation of the tangent line.
1. Find the derivative of the given function:
y = x^3 + 3x^2 + 2
Taking the derivative with respect to x:
dy/dx = 3x^2 + 6x
2. Set the derivative equal to zero and solve for x:
3x^2 + 6x = 0
Factor out common factor of 3x:
3x(x + 2) = 0
Set each factor equal to zero:
x = 0 or x = -2
3. Substitute the x-coordinate into the original function to find the y-coordinate:
For x = 0:
y = (0)^3 + 3(0)^2 + 2 = 2
For x = -2:
y = (-2)^3 + 3(-2)^2 + 2 = -2
Therefore, the point of inflection is (0, 2) and (-2, -2).
4. Use the point-slope form of a line to find the equation of the tangent line:
The slope of the tangent line is the derivative of the function at the given point of inflection.
Using the derivative calculated earlier, when x = 0:
slope = dy/dx = 3(0)^2 + 6(0) = 0
Using the point-slope form: y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
For (x1, y1) = (0, 2):
y - 2 = 0(x - 0)
y - 2 = 0
y = 2
Therefore, the equation of the line tangent to y = x^3 + 3x^2 + 2 at its point of inflection is y = 2.