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anonymous
 4 years ago
THE EQUATION OF ACURVE IS Y+2X+8/X^2
iii)Show that the normal to the curve at the point (2,2) intersects the x axis at the point (10,0)
anonymous
 4 years ago
THE EQUATION OF ACURVE IS Y+2X+8/X^2 iii)Show that the normal to the curve at the point (2,2) intersects the x axis at the point (10,0)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sry guys the equation is y=2x+8/x^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(2x+8)/x^2\] Like this?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the normal might be the second derivative if i remember this correctly

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.14 (x+12)/x^4 is what the wolf spits out; see if that works for the slope of the line between the 2 points to test

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wasnt the normal the negative reciprocal of the first derivative?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.11/4 is the slope between the points: 4(2+12)/16 = 10/4 = 5/2 bummer yeah, the perp slope lol

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i musta been doing calc3 vectors

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1(2x+8)/x^3; x=2 (4+8)/8 = 1/4 so slope is 4 perhaps?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.110,0 2 , 2  8,2; slope = 1/4 hmmm

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the slope is 1/4 u good :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i cant get the slopes to match up, we sure there aint a typo, in yours or mine?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1other than 1/4 spose to be 1/2 the perps to 2 :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.12 + 8/x^2 downs to: 0 16/x^3 at x=2 we get 16/8 = 2, perps to 1/2 is locser

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1forgot an "x" lol 2x + 8x^2; 2  16x3; 2 + 2 = 4; which perps tp 1/4 got it lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don`t get it how did we arrive to 1/4?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1equation; first derivative = slope at any point; normal is perp slope; perp slope of m = 1/m

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0amistre=amazing XD could you explain how you did it to me too :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1the eq was:\[2x+8x^{2}\]to begin with

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1once you determine the perp sloped derivative :) compare it with the slope of the line between the given points

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1thnx, and good luck ;)
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