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

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

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

There is an explanation video available below.

{{ settings.no_comment_msg ? settings.no_comment_msg : 'There are no comments' }}