+0  
 
0
111
5
avatar+792 

A curve has equation 𝑦 = 1/𝑥 + 4𝑥 .

 The point P(-1, -5) lies on the curve. Find an equation of the Normal to the curve at point P.

i used dy/dx to get -x^-2 +4 

then i subbed in -1 which gets me 3, 3 is the gradient for the tangent 

so m*m = -1 m=-3

then do i just sub the information in y-y1 =m(x-x1) ?

 Dec 2, 2018
edited by YEEEEEET  Dec 2, 2018

Best Answer 

 #2
avatar+17332 
+2

What is the equation of the curve ?    1x + 4x = 5x

    I think you ment   1/x + 4x

    x^-1 +4x     the derivative will tell you the slope at a given point

Derivative

-x^-2  + 4        at x = -1    yields  -1 +4 = 3 = slope      Normal (perpindicular ) slope will be   - 1/3

 

y = -1/3 x +b    contains the point    -1,-5    sub that in

-5 = -1/3 (-1) + b

-5 = 1/3 +b     b = -5 1/3

 

y = -1/3 x  - 16/3 

 Dec 2, 2018
 #1
avatar+98196 
+1

What is the equation of the curve???

y = 1x + 4x  = 5x       has a derivative of 5

 

 

cool cool cool

 Dec 2, 2018
 #3
avatar+792 
+1

sorry my bad i typed it wrong it shouldve been 1/x

YEEEEEET  Dec 2, 2018
 #2
avatar+17332 
+2
Best Answer

What is the equation of the curve ?    1x + 4x = 5x

    I think you ment   1/x + 4x

    x^-1 +4x     the derivative will tell you the slope at a given point

Derivative

-x^-2  + 4        at x = -1    yields  -1 +4 = 3 = slope      Normal (perpindicular ) slope will be   - 1/3

 

y = -1/3 x +b    contains the point    -1,-5    sub that in

-5 = -1/3 (-1) + b

-5 = 1/3 +b     b = -5 1/3

 

y = -1/3 x  - 16/3 

ElectricPavlov Dec 2, 2018
 #4
avatar+792 
+1

thank you!

YEEEEEET  Dec 2, 2018
 #5
avatar+98196 
0

Thanks, EP.....

 

 

cool cool cool

CPhill  Dec 2, 2018

14 Online Users

avatar
avatar