Question

The gradient of the curve is given by dy/dx=ax+b/x^2. If the curve passes through the point (-1,-4) and has a turning point at the point (1,0) find its equation and sketch its curve.

i want to know how to get to the answer. the answer to equation is y=x^2+2/x-3.

Answer 1

i have integrated the gradient to get y=ax^2/2-b/x+c

Answer 2

Ok perfect, so now what happens when you try to use (-1,-4) and (1,0) to solve for these constants?

Answer 3

ok do i plug in the first set of points (-4)=a/2+b
the second set of points 0=a/2+b

Answer 4

and then i solve as simultaneous equations for a and b is that correct?

Answer 5

Wait, what happened to c?

Answer 6

We have three unknowns a,b, and c to solve for so we need 3 equations.

The third equation you will need comes from understanding how to find the turning point at (1,0) with the derivative.

Answer 7

can you summarise how we do that again as am slightly confused.

Answer 8

Yeah sure, so remember the derivative is the slope of the function you are solving for the constants a, b, and c right? So since they say that (1,0) is a turning point that means that the derivative at this point is 0. For example, at the bottom of a parabola, the slope is 0 because that's where it turns around to go back up.

So using this:
\[\Large \frac{dy}{dx}=ax+\frac{b}{x^2}\]

So here we have x=1, y=0, and dy/dx=0. If this was an implicit derivative we'd be able to plug in y, but it doesn't show up here. Plugging in we get:

\[\Large 0=a1+\frac{b}{1^2}\]

Now you can use this equation along with the two you had earlier (don't forget the c of integration!) to solve it as a system of equations like you describe.

Answer 9

thanks buddy i was having a bath as my computer froze thanks you have been very helpful and i really appreciate that/

Answer 10

Yeah glad I could help!