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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

I've been working out the gradient (and whole equation) of the tangent of certain points on a curve but this new set of questions wants the equation of the normal. I figure I'd just work as I had been and get the equation of the tangent, then simply flip over the gradient and reverse the sign for the normal i.e.

tangent; y = 7x - 13

normal; y = -1/7x - 13

The full sum is as follows:

y = 2x^2 - 5x + 3 at (2, 1)

dy/dx = 6x - 5

12 - 5 = 7

(y - 1)/(x - 2) = 7

y - 1 = 7(x - 2) = 7x - 14

y = 7x - 13

and thus the normal; y = -1/7x - 13.

The book claims x + 3y = 5

Aloha Nui means Goodbye.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

dy/dx = 6x - 5

dy/dx = 4x - 5, slope at (1, 2) is 3

Try it now.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

Aloha Nui means Goodbye.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Don't worry about it, on a math test, I wrote:

2x = 2

x = 2

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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