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) ?

YEEEEEET Dec 2, 2018

#2**0 **

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

#2**0 **

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