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Equations of tangents to curves

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

\frac{4}{x^2} = 4x^{-2}

(Remember that

x^{-n} = \frac{1}{x^n}

)

(edited 13 years ago)

Reply 3

im so academic

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.

(edited 13 years ago)

Reply 4

Farhan.Hanif93

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.

Reply 5

boromir9111

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

Reply 6

Pharm Boy

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. :smile:

Reply 7

Pharm Boy

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. :wink:

Reply 8

im so academic

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.