Posted by **Anonymous** on Sunday, November 18, 2007 at 12:49am.

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).

Work...

4(1)-3=1

y-1=1(x-1)

y=x

- calculus -
**Reiny**, Sunday, November 18, 2007 at 1:26am
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 -
**Anonymous**, Sunday, November 18, 2007 at 11:43am
why would I solve for the constant? I get k=-4. What is meant by curve?

- calculus -
**Reiny**, Sunday, November 18, 2007 at 3:01pm
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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