Find the equation of the tangent at the point (2, 0) to the curve y = x\(^2\) - 2x

A.

y = 2x - 4

B.

y = 2x + 4

C.

y = 2x - 2

D.

y = 2x + 2

Correct answer is A

The gradient to the curve is found by differentiating the curve equation with respect to x

  So \(\frac{dy}{dx}\) 2x - 2

  The gradient of the curve is the same with that of the tangent.

  At point (2, 0) \(\frac{dy}{dx}\) = 2(2) - 2

  = 4 – 2 = 2

  The equation of the tangent is given by (y - y1) \(\frac{dy}{dx}\) (x – x1)

  At point (x1, y1) = (2, 0)

  y - 0 = 2(x - 2)

  y = 2x - 4