If the graph of y=f(x) contains the point (0,-5) and if dy/dx=x-3/y, then what is f(x)?

Answer 1

Assuming that the problem is dy/dx = (x - 3)/y:

To make it a little simpler, we can multiply both sides by 2 to get:

y² = x² - 6x + C₂

We are told this is a function, so we can take the square root but given that (0, -5) is a point, this MUST be the negative square root so:

y = -√(x² - 6x + C₂)

Plugging in (0, -5), we get:

-5 = -√((0)² - 6(0) + C₂)

-5 = -√C₂ therefore C₂ = 25

So f(x) = -√(x² - 6x + 25)