Question

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)

Answer 1

sry guys the equation is y=2x+8/x^2

Answer 2

\[(2x+8)/x^2\]
Like this?

Answer 3

yes

Answer 4

the normal might be the second derivative if i remember this correctly

Answer 5

4 (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

Answer 6

Wasnt the normal the negative reciprocal of the first derivative?

Answer 7

-1/4 is the slope between the points:

4(-2+12)/16 = 10/4 = 5/2

bummer

yeah, the perp slope lol

Answer 8

i musta been doing calc3 vectors

Answer 9

XD

Answer 10

-(2x+8)/x^3; x=-2

-(-4+8)/-8 = 1/4

so slope is -4 perhaps?

Answer 11

-10,0
2 , 2
-------
-8,2; slope = -1/4 hmmm

Answer 12

the slope is -1/4 u good :)

Answer 13

i cant get the slopes to match up, we sure there aint a typo, in yours or mine?

Answer 14

other than 1/4 spose to be 1/2 the perps to -2 :)

Answer 15

2 + 8/x^2

downs to: 0 -16/x^3

at x=-2 we get -16/-8 = 2, perps to -1/2 is locser

Answer 16

forgot an "x" lol

2x + 8x^-2; 2 - 16x-3; 2 + 2 = 4; which perps tp -1/4

got it lol

Answer 17

i don`t get it how did we arrive to -1/4?

Answer 18

which part?

Answer 19

equation; first derivative = slope at any point; normal is perp slope; perp slope of m = -1/m

Answer 20

amistre=amazing XD could you explain how you did it to me too :)

Answer 21

the eq was:\[2x+8x^{-2}\]to begin with

Answer 22

once you determine the perp sloped derivative :) compare it with the slope of the line between the given points

Answer 23

thats smart :)

Answer 24

thnx, and good luck ;)

Answer 25

thank u next tym :)