The following curve passes through (3,1). Use the local linearization of the curve to find the approximate value of y at x=2.8 ...?

d/dx (2 x^2 y + y = 2x + 13) 4xy + 2x^2 y' + y' = 2 4xy + y' (2x^2 + 1) = 2 y' = (2 - 4xy) / (2x^2 + 1)

ow we can use this in a linear equation for a slope Ty = - 5x/8 + 5 (3) / 8 + 8/8 = - 5x/8 + (15+8) / 8 = - 5x/8 + 23/8 this will gives us an approximation at x=2.8 now