Thread: the point tangent to line

At what point on the curve xy = ( 1 - x - y)^2 is the tangent line parallel to the x - axis


need your assistance here guys...

Originally Posted by rcs

At what point on the curve xy = ( 1 - x - y)^2 is the tangent line parallel to the x - axis


need your assistance here guys...

What is the gradient of the x axis? Hint: It's horizontal...

Sir that term i coudnt answer ... im sorry i couldnt get the hint

Originally Posted by rcs

At what point on the curve xy = ( 1 - x - y)^2 is the tangent line parallel to the x - axis

now find a point so that .

If you do not know what "gradient' (American English: "slope") of a line is, you should not be attempting a problem like this. Finding the derivative is far more advanced that the 'slope' or 'gradient' of a line.

Originally Posted by rcs

At what point on the curve xy = ( 1 - x - y)^2 is the tangent line parallel to the x - axis


need your assistance here guys...

You've posted a lot of algebra questions on this forum, and this question might be too advanced for you. How much calculus do you know?

i guess basic for u but im trying to know all these things sir

Originally Posted by rcs

i guess basic for u but im trying to know all these things sir

What the hell does that mean?
That is not even a well formed English sentence
If you cannot post an proper statement or question then post nothing !

Originally Posted by rcs

i guess basic for u but im trying to know all these things sir

I don't think you're quite ready for derivatives and calculus, especially since you've been posting elementary algebra questions. You need to be proficient in algebra to do well in calculus.

The solution would involve differentiating both sides with respect to x:

(using chain rule)

Then set dy/dx = 0.

thank you Plato. You have a very positive statement . Keep it up!