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# don't know how to do it.

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A curve C has equation 𝑦=𝑥^3 −𝑥^2 −𝑥+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
+2

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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yes i got it thank you so much

Guest Jan 17, 2021
#2
+11086
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A curve C has equation 𝑦=𝑥^3 −𝑥^2 −𝑥+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)

$$𝑦=𝑥^3 −𝑥^2 −𝑥+2\\ \color{blue}\frac{dy}{dx}=3x^2-2x-1$$

b)

$$P(2,y(2))\\ P(2,2^3 −2^2 −2+2)$$

$$P(2,4)$$

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

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

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

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)

!

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