Math help (1 Viewer)

icycledough

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The normal will be perpendicular to the tangent at x = 0. So You should first differentiate the equation and sub in x = 0 to find the gradient of the tangent at x = 0. Then, knowing that m1 * m2 = -1 (where m1 and m2 are the gradients of the tangent and the normal respectively), you can find m2 by rearranging the equation. Then, you will have the gradient of the normal and the co-ordinates when x = 0 (which I believe is 0,0 as the numerator is 0 and the denominator is 1). Then, you can apply the point-gradient formula to get the equation
 

specificagent1

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The normal will be perpendicular to the tangent at x = 0. So You should first differentiate the equation and sub in x = 0 to find the gradient of the tangent at x = 0. Then, knowing that m1 * m2 = -1 (where m1 and m2 are the gradients of the tangent and the normal respectively), you can find m2 by rearranging the equation. Then, you will have the gradient of the normal and the co-ordinates when x = 0 (which I believe is 0,0 as the numerator is 0 and the denominator is 1). Then, you can apply the point-gradient formula to get the equation
The issue is when i differentiate it and sub x = 0 the gradient is 0 and hence breaks the m1 x m2 formula
 

specificagent1

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I'm not sure if you have the answer for this question or not, but I think it might be x=0 (a vertical line). But again, I'm not 100% sure on it
I dont have the answers but graphically x = 0 makes sense but i was just wondering how you would reach that via the algebra
 

icycledough

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I dont have the answers but graphically x = 0 makes sense but i was just wondering how you would reach that via the algebra
Yh, I typed it into Desmos and just intuitively thought that it may be the normal (as I thought y = 0 would be the tangent). But yh, unless you were able to graph it, it may be difficult to get it through algebra (calculus means)
 

CM_Tutor

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If the gradient of the tangent is 0 then the normal must be vertical and have no defined gradient, and the form of the normal must be for some constant .
 

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