Hi,

Find the equation of the tangent to the curve 2x2 + y - xy2 - 2y3 = 4 at the point (-1, -1).

I am confused because substituting the x and y coordinates into the above equation does not equal to 4. Does this simply mean that the line is not on the curve?

I solved for dy/dx and got (-4x - y2) / (1-x2y-6y3) and substituting (-1, -1) into this equation I got the slope of the tangent to the curve to be y= 1/3 (x+1) -1.

Does this seem right?

Thank you

Hi,

Find the equation of the tangent to the curve 2x2 + y - xy2 - 2y3 = 4 at the point (-1, -1).

I am confused because substituting the x and y coordinates into the above equation does not equal to 4. Does this simply mean that the line is not on the curve?

I solved for dy/dx and got (-4x - y2) / (1-x2y-6y3) and substituting (-1, -1) into this equation I got the slope of the tangent to the curve to be y= 1/3 (x+1) -1.

Does this seem right?

Thank you
check your math. (-1,-1) is on that curve.

your derivative isn't quite correct but it's close enough that I believe you have the right idea. Check your algebra and differentiation.

Once you have the proper derivative formula just evaluate it at (-1,-1) to get a value for the slope at that point.

Then with the point (-1,-1) and the slope you can produce the equation for the tangent line using point slope form.