Determine the equation ofthetangent line in slopeintercept form at x=1 of y^3 +2x^2=1 at x=1. Please show work so I can follow the process. Thanks!

Answer 1

y^3+2x^2=1

y^2 dy/dx+4x=0
dy/dx=-4x/y^2

at x=1, y=-1, so

dy/dx=-4/1=-4
check my thinking.

Answer 2

Hmm. I get

y^3 + 2x^2 = 1
3y^2 y' + 4x = 0
y' = -4x/(3y^2)

y(1) = -1, so
y'(1) = -4/3

so, now you have a point (1,-1) and a slope: -4/3

y+1 = -4/3 (x-1)

I assume that massaging that to slope-intercept form presents no problems...