Respuesta :
The equation of the line tangent to the given line is given by: y = -(3x-1).
What is the equation of a line tangent to another line?
The equation of the tangent line can be found using the formula:
[tex]y-y_{1} = m(x-x_{1})[/tex], where m is the slope and ([tex]x_{1},y_{1}[/tex]) is the coordinate points of the line.
Given: The equation of line: [tex]y = x^{3}+3x^{2} +2 \\[/tex]
The first and second derivatives of the equation will be given by:
[tex]= > y' = 3x^{2} +6x[/tex]
[tex]= > y''=6x+6[/tex]
- As the point of inflection is the point at the curve where the second derivative = 0, here the point of inflection:
POI => 6x + 6 = 0
=> x = -1
- Now, for x = -1, [tex]y'=3\times(-1)^{2}+6(-1) = -3[/tex]
(y' = slope of the equation; as slope of a line = slope of a line tangent to it)
- Next, for x = -1 , y = [tex](-1)^{3} + 3(-1)^{2} +2= -1+3+2 =4[/tex]
Hence, the equation of the line, tangent to given line:
y - 4 = (-3) ( x + 1)
=> y = - (3x - 1)
To learn more about equations of lines, refer to the link: https://brainly.com/question/6617153
