Find the equation of the tangent at the point (2, 0) to the curve y = x2 2 - 2x

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

Answer Details

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