Original post by slothgal
Please can someone help, ive tried to do this so many times
1)find the x coordinate of the point on the curve y= (x^3)/3 - 3x^2 + 2/3 where the tangent is parallel to the tangent drawn at point (1,2)
thank you
1)find the x coordinate of the point on the curve y= (x^3)/3 - 3x^2 + 2/3 where the tangent is parallel to the tangent drawn at point (1,2)
thank you
The point (1,2) isn't on the curve. Is that intentional, or should it be (1,-2) ?
Original post by slothgal
bump please help
unpick the question
You want 2 parallel lines
You will need their gradient
How will you find the gradient of the tangent to the curve at (1, -2)
Original post by slothgal
parallel lines have the same gradient, so it has the same gradient as the curve?
yes
Original post by slothgal
sorry how can i differentiate (x^3)/3
How would you usually differentiate
Original post by slothgal
x^3 is 3x^2
What you're trying to say is that the first derivative of with respect to is .
Original post by slothgal
x^3 is 3x^2
yep
So if you divide that by 3 you will get ... ...
Original post by slothgal
X^2! Thank you, im a fool
less of a fool if you use the quote facility so that I can see you have responded
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