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

Respuesta :

lucic

Answer:

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

Step-by-step explanation:

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

Find dy/dx

[tex]\frac{d}{dx} (y)=\frac{d}{dx}(2x^3)[/tex]

Apply the power rule

[tex]\frac{d}{dx} (x^n)=nx^{n-1}[/tex]

where n=constant

Hence, our equation will be;

[tex]\frac{dy}{dx} =2*3x^{3-1} \\\\\\\frac{dy}{dx} =6x^{2}[/tex]

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)