Find the equation of the tangent at the point (2, 0) to the curve y = x2 - 2x
A.
y = 2x - 4
B.
y = 2x + 4
C.
y = 2x - 2
D.
y = 2x + 2
Correct Answer: y = 2x - 4
Explanation
The gradient to the curve is found by differentiating the curve equation with respect to x
So dy/dx 2x - 2
The gradient of the curve is the same with that of the tangent.
At point (2, 0) dy/dx = 2(2) - 2
= 4 – 2 = 2
The equation of the tangent is given by (y - y1) 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 dy/dx 2x - 2
The gradient of the curve is the same with that of the tangent.
At point (2, 0) dy/dx = 2(2) - 2
= 4 – 2 = 2
The equation of the tangent is given by (y - y1) dy/dx
(x – x1)
At point (x1, y1) = (2, 0)
y - 0 = 2(x - 2)
y = 2x - 4
