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# A curve has equation 𝑦 = 1 𝑥 + 4𝑥 .

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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

#2
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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
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What is the equation of the curve???

y = 1x + 4x  = 5x       has a derivative of 5   Dec 2, 2018
#3
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sorry my bad i typed it wrong it shouldve been 1/x

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
#4
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thank you!

YEEEEEET  Dec 2, 2018
#5
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Thanks, EP.....   CPhill  Dec 2, 2018