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
Verified Correct Answer: A.
Step-by-Step Mathematical Solution
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
Source: Myschool Past Questions Archive (Record #47566)