Find the equation of the tangent at the point (2, 0) to the curve y…

Explanation

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

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