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The slope of a curve is at the point (x,y) is 4x-3. Find the curve if it is required to pass through the point (1,1).


  • calculus - ,

    4x - 3 is your slope, so it is the derivative.

    then dy/dx = 4x + 3, integrate to get

    y = 2x^2 + 3x + k

    plug in (1,1) into that to get k, and you are done!

  • calculus - ,

    why would I solve for the constant? I get k=-4. What is meant by curve?

  • calculus - ,

    When they say "find the curve" they mean find the equation of function whose graph would be that curve.
    your equation y = 2x^2 + 3x - 4 would graph to be a parabola, and a parabola is a curve.

    As to the constant, remember that if you differentiate an equation like

    y = 2x^2 + 3x - 4
    you get y' = 4x + 3 - 0

    so when we "anti-differentiate" that we really don't know what the value of the constant was, because its derivative would be zero no matter what the number was.
    That is why we include a constant value of c or k to allow for that. Once we sub in the given point we then know the value of that constant

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