Find the points on the curve y= x^3 -6 where the tangents are parallel to the line y=12x-1

Hence find the equations of the normals to the curve at those points.


Please help! I don't know how to approach this question please someone tell me what to do !

first of all, work out the gradient of the line.
y = 12x - 1
therefore the gradient is 12

You want to find points on y = x^3 - 6 which have a gradient of 12. To find this, you differentiate the equation.

y = x^3 - 6
dy/dx = 3x^2

Because the differential equations gives the gradient at any point x, by substituting "12" into dy/dx you can work out any x values with a gradient of 12.

12 = 3x^2
4 = x^2
x = 2
x = -2

Now to work out the coordinates you substitute these values of x into the original equation:

y = x^3 - 6
y = (2)^3 - 6
y = 2
Your first point: (2, 2)
y = x^3 - 6
y = (-2)^3 - 6
y = -14
Your second point: (-2, -14)

The gradient of a normal is the negative inverse of the gradient of the tangent: That is, it will be equal to -1/12. You can find the equations of the normals using gradient form equations.

y = mx + c
2 = -1/12(2) + c
c = 2,1/6
Normal one: y = -1/12x + 2,1/6

y = mx + c
-14 = -1/12(-2) + c
c = -14,1/6
Normal two: y = -1/12x - 14,1/6

Derive y = x^3 - 6 and find when dy/dx = 12

y = x^3 - 6
dy/dx = 3x^2
dy/dx = 12
12 = 3x^2
4 = x^2
-2 , 2 = x

Now, find the y-values for these

y = x^3 - 6
y = -8 - 6 , 8 - 6
y = -14 , 2

Now, describe normal lines that pass through the coordinates

y - k = m * (x - h)

m = -1/12

y - (-14) = (-1/12) * (x - (-2))
y + 14 = (-1/12) * (x + 2)
-12y - 168 = x + 2
-12y = x + 170
y = (-1/12) * (x + 170)

y - 2 = (-1/12) * (x - 2)
-12y + 24 = x - 2
-12y = x - 26
y = (-1/12) * (x - 26)

If the tangents are parallel to the given line the derivative will equal the line's slope, so:

y' = 3x^2 = 12

solve the equation to get x = 2 and -2
plug back into y=x^3-6 to get coordinates

(2, 2) and (-2, -14)