A calculus problem by A Former Brilliant Member
Calculus
Level
2
The tangent to the curve y=2x^2-x+1 is parallel to the line y= 3x+9 at the point ..........
(2,1)
(1,2)
(3,9)
(3,7)
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1 solution
As the tangent is parallel to the line with equation y = 3x + 9 so, the gradient of the tangent will be 3.
dy/dx of the curve = 3 as dy/dx represent the gradient.... therefore, 3 = 4x - 1 (d/dx of 2x^2 - x + 1)
x = 1 and solving for y we get y = 2 so the coordinate is (1,2)