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Question
The equation of tangent to the curve y = 3x2 - x + 1 at the point (1, 3) is
(a) y=5x+2
(b)y=5x-2
(c)y=1/5x+2
(d)y=1/5x-2
Solution
y = 5x - 2
`dy/dx=6x-1" at "(1,3)`
Slope of the tangent at (1, 3) = (6 - 1) = 5
Equation of tangent is y - y1 = m(x - x1)
y - 3 = 5(x - 1)
5x - y - 2 = 0
y = 5x - 2
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