Question

How to find the coordinates of all points on the curve 2x^3 which the tangent line has slope 6?

Answer 1

(1,2) (-1,-2)

Explanation:

Given the equation of curve as y= 2x³ and slope of tangent line as 6 then

Find dy/dx

`(d)/(dx) (y)=(d)/(dx)(2x^3)`

Apply the power rule

`(d)/(dx) (x^n)=nx^(n-1)`

where n=constant

Hence, our equation will be;

`(dy)/(dx) =2*3x^(3-1) \\\\\\(dy)/(dx) =6x^(2)`

But you know that dy/dx=slope=6

6x²=6--------------------divide both sides by 6

6x²/6=6/6

x²=1

x=√1=±1

x=1 and -1

Remember y=2x³

Substitute value of x to get value of y

y=2x³

y=2×1³

y=2×1=2

y=2

For x=-1, find y coordinate

y=2×-1³=2×-1=-2

coordinate will be (-1,-2)

Coordinates of the points will be (1,2) ,(-1,-2)