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Calculus, 2ADV C4 2011 HSC 4c

The gradient of a curve is given by  `dy/dx = 6x-2`.  The curve passes through the point `(-1, 4)`. 

What is the equation of the curve?   (2 marks)

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 `y = 3x^2-2x-1`

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`dy/dx = 6x-2` 

`y` `= int 6x-2\ dx`
  `= 3x^2-2x + c`

 
`text{Since it passes through}\ (-1,4),`

`4` `= 3 (-1)^2-2(-1) + c`
`4` `= 3 + 2 + c`
`c` `= -1`

 
`:. y = 3x^2-2x-1`

Filed Under: Integrals, Other Integration Applications (Y12), Tangents and Normals Tagged With: Band 4, smc-1089-30-Find f(x) given f'(x), smc-1090-10-Quadratic Function, smc-1090-50-Find curve given tangent, smc-1213-25-Tangents/Primitive function

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