Find the equations of the tangents to y = x^3 from the external point (-2, 0).

Question

Find the equations of the tangents to y = x^3 from the external point (-2, 0).

 

Answer 1

Let P1 (-2,0) be a point ON THE TANGENT LINE, BUT NOT THE POINT OF TANGENCY.

 

At a general point ON THE CUBIC curve, the slope of the tangent is m = 3x².

 

Using point slope form for the equation of a line passing through P1:  y - 0 = m(x - (-2))

y = 3x² (x+2)

 

At point of tangency T(x,x³) and substituting its coordinates into the above line's equation:

x³ = 3x²(x+2) 

0 = 2x³+6x²

0 = 2x²(3+x)  which gives solution x = 0 or -3 as the POINTS OF TANGENCY.

 

Now use two cases above to find the associated slope values and y coordinates of POINTS OF TANGENCY, and finally write the two equations of the two tangents the pass through OTHER POINT (-2,0).

 

Answer 2

Take the derivative of the function.  This will be the slope of the tangent line.

 

m = dy/dx

 

m = 3x²

 

m = 3(-2)²

 

 

Then use the point-slope form of a line.

 

 

y = m(x - x₁) + y₁

 

where:

m = slope

(x₁ , y₂) is the given point