Determine whether or not F is a conservative vector field. If it is, find a function f such that F = ∇f. (If the vector field is not conservative, enter DNE.) F(x, y) = (y2 − 4x)i + 2xyj

f = xy²-2x² satisfies that  F = ∇f.

Step-by-step explanation:

F(x,y) = (y² - 4x) i + 2xy j

We want f(x,y) such that

Lets find a primitive of y²-4x respect to the variable x. We need to think y (and y²) as constants here, so a primitive of y² would be xy² the same way that a primitive of k is xk (because we treat y² as constant). A primitive of x is x²/2, thus a primitive of 4x is 2x². Thus, a primitive of y²-4x is xy² - 2x². We can obtain any other primitive by summing a constant, however since we treated y as constant, then we have that

where c(y) only depends on y (thus, it is constant repsect with x).

We will derivate the expression in terms of y to obtain information about c(y)

Thus, is constant. We can take f(x,y) = xy²-2x². This function f satisfies that  F = ∇f.

both equations and proportions are better.

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.