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