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A curve C has equation 𝑦=π‘₯^3 βˆ’π‘₯^2 βˆ’π‘₯+2

The point P has x-coordinate 2

 

a) Find dy/dx in terms of x.

 b) Find the equation of the tangent to the curve C at the point P. The normal to C at P intersects the x-axis at A.

c) Find the coordinates of A.

 Jan 17, 2021
 #1
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A.  the first derivative will be   3 x2  - 2x  - 1

B .   find the 'y' coordinate of point 'P' by sub'ing x = 2 into the original equation

          then find the SLOPE at point 'P' by sub'ing in x = 2 into the first derivative equation found in A.

       use this slope to calculate the line equation given point 'P'

         perpindicular to  this slope is   - 1 / (this slope)      

               can you take it from here?   

 Jan 17, 2021
 #3
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+2

yes i got it thank you so much 

Guest Jan 17, 2021
 #2
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+1

A curve C has equation 𝑦=π‘₯^3 βˆ’π‘₯^2 βˆ’π‘₯+2

The point P has x-coordinate 2

 

a) Find dy/dx in terms of x.

b) Find the equation of the tangent to the curve C at the point P.

   The normal to C at P intersects the x-axis at A.

c) Find the coordinates of A.

 

Hello Guest!

 

a)

\(𝑦=π‘₯^3 βˆ’π‘₯^2 βˆ’π‘₯+2\\ \color{blue}\frac{dy}{dx}=3x^2-2x-1\)

 

b)

\(P(2,y(2))\\ P(2,2^3 βˆ’2^2 βˆ’2+2)\)

\(P(2,4)\)

\(y'=3x^2-2x-1\\ m=y'(2)\\ m=3\cdot 2^2-2\cdot 2-1\\ \color{blue}m=7\\ E_P(x)=m(x-x_P)+y_P\)

\(E_P(x)=7(x-2)+4\\ =7x-14+4\)

\(E_P(x)=7x-10\)

 

c)

\(N_{PA}(x)=-\frac{1}{m}(x-x_P)+y_P\\ N_{PA}(x)=-\frac{1}{7}(x-2)+4\\ =-\frac{1}{7}x+\frac{2}{7}+4\)

\(N_{PA}(x)=-\frac{1}{7}x+\frac{30}{7}\)

\(-\frac{1}{7}x+\frac{30}{7}=0\\ -\frac{1}{7}x=-\frac{30}{7} \)

\(x=30\)

\(A(30,0)\)

 

The coordinates of A are A(30,0)

laugh  !

 Jan 17, 2021
edited by asinus  Jan 17, 2021
edited by asinus  Jan 17, 2021
edited by asinus  Jan 17, 2021

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