A curve C has equation π¦=π₯^3 βπ₯^2 βπ₯+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.
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?
A curve C has equation π¦=π₯^3 βπ₯^2 βπ₯+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)
\(π¦=π₯^3 βπ₯^2 βπ₯+2\\ \color{blue}\frac{dy}{dx}=3x^2-2x-1\)
b)
\(P(2,y(2))\\ P(2,2^3 β2^2 β2+2)\)
\(P(2,4)\)
\(y'=3x^2-2x-1\\ m=y'(2)\\ m=3\cdot 2^2-2\cdot 2-1\\ \color{blue}m=7\\ E_P(x)=m(x-x_P)+y_P\)
\(E_P(x)=7(x-2)+4\\ =7x-14+4\)
\(E_P(x)=7x-10\)
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)
!