Question

find the slope of the tan line at the specified point.
y^3+yx^2+x2-3y^2=0; (0,3)

Answer 1

i got that dy/dx=(-2xy-2x)/3y^2+x^2-6y. but what do I do with the (0,3)?

Answer 2

Merely plug in x and y into the derivative and you get the slope.

Answer 3

so it would be 0?

Answer 4

It would be -18, because the terms with x's cancel out.

Answer 5

wouldnt that make the top 0?