+0  
 
0
71
5
avatar+639 

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+15350 
+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+94545 
+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+639 
+1

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

YEEEEEET  Dec 2, 2018
 #2
avatar+15350 
+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+639 
+1

thank you!

YEEEEEET  Dec 2, 2018
 #5
avatar+94545 
0

Thanks, EP.....

 

 

cool cool cool

CPhill  Dec 2, 2018

9 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.