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# Equations of tangents to curves Watch

1. I've got the question:

y=4/x^2 at x=-1

I would be able to do it, but the fraction threw me off. Little help please?
2. y = 4*x^-2
3. (Remember that )
4. (Original post by JoshCoulthard)
I've got the question:

y=4/x^2 at x=-1

I would be able to do it, but the fraction threw me off. Little help please?
y= 4/x^2 is the same thing as y=4x^-2

First find the coordinate of the point (sub x = -1 into the equation to find y)
Then dy/dx to find the gradient.
Then use y-y1=m(x-x1) to find the equation.
5. (Original post by im so academic)
Remember that you are asked to find the tangent, so it's -1/m.
I'm afraid you're wrong there. If m is the value of dy/dx at the point with x=-1 then -1/m will give you the gradient of the normal to the curve at that point and not the tangent.
6. (Original post by Farhan.Hanif93)
I'm afraid you're wrong there. If m is the value of dy/dx at the point with x=-1 then -1/m will give you the gradient of the normal to the curve at that point and not the tangent.
Beat me to it lol
7. Yeah you just fling the bottom part of the fraction to the top (which means it's now multiplying the top number) and change the symbol of the power.
8. (Original post by im so academic)
y= 4/x^2 is the same thing as y=4x^-2

First find the coordinate of the point (sub x = -1 into the equation to find y)
Then dy/dx to find the gradient. Remember that you are asked to find the tangent, so it's -1/m.
Then use y-y1=m(x-x1) to find the equation.
How academic of you.
9. (Original post by Farhan.Hanif93)
I'm afraid you're wrong there. If m is the value of dy/dx at the point with x=-1 then -1/m will give you the gradient of the normal to the curve at that point and not the tangent.
Damn it, I knew you didn't have to do the -1/m thing, but never mind.

Tangent = don't edit gradient
Normal/perpendicular = edit gradient

Utter fail. Thank you though.

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Updated: December 29, 2010
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