Answer:
[tex]y = -3x +1[/tex]
Explanation:
Given
[tex]y = x^3 + 3x^2 + 2[/tex]
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:
[tex]y = x^3 + 3x^2 + 2[/tex]
First derivative is:
[tex]y' = 3x^2 + 6x[/tex]
Second derivative:
[tex]y" = 6x + 6[/tex]
Equate to 0
[tex]6x + 6 = 0[/tex]
Solve for x
[tex]6x = -6[/tex]
[tex]x = -1[/tex]
Next, we calculate the slope (m) of the point using the first derivative.
Substitute -1 for x in [tex]y' = 3x^2 + 6x[/tex]
[tex]m = 3*-1^2 + 6 * -1[/tex]
[tex]m = 3*1 - 6[/tex]
[tex]m = 3 - 6[/tex]
[tex]m = -3[/tex]
To get the y equivalent;
Substitute [tex]x = -1[/tex] in [tex]y = x^3 + 3x^2 + 2[/tex]
[tex]y = (-1)^3 + 3(-1)^2 + 2[/tex]
[tex]y = -1 + 3+ 2[/tex]
[tex]y = 4[/tex]
Lastly, we calculate the line equation using:
[tex]y - y_1 =m(x- x_1)[/tex]
Where
[tex]x = -1[/tex] and [tex]y = 4[/tex]
[tex]m = -3[/tex]
So, we have:
[tex]y - 4 = -3(x -(-1))[/tex]
[tex]y - 4 = -3(x +1)[/tex]
[tex]y - 4 = -3x -3[/tex]
Make y the subject
[tex]y = -3x -3 +4[/tex]
[tex]y = -3x +1[/tex]
Hence, the equation is: [tex]y = -3x +1[/tex]
