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# C1 Maths question help watch

1. I've tried this question out and I've done a and b but can't do part c.

The curve C has equation y=x^3 - 2x^2 - x + 9 x>0

The point P has coordinates (2,7)

a) Show that P lies on C

worked out to be ;
7= 2^3 -2(2)^2-2+9
7 = 8-8-2+9
7=7

b) Find the equation of the tangent to C at P, giving your answer in the form y=mx+c, where m and c are constants.

found to be ;
dy/dx= 3x^2-4x - 1
12-8-1
m = 3
y-7=3(x-2)
y = 3x+1
The point Q also lies on C

Given that the tangent to C at Q is perpendicular to the tangent to C at P,

c) show that the x coordinates of Q is 1/3 (2 + square root 6)
??
2. (Original post by shaunobrien1)
I've tried this question out and I've done a and b but can't do part c.

The curve C has equation y=x^3 - 2x^2 - x + 9 x>0

The point P has coordinates (2,7)

a) Show that P lies on C

worked out to be ;
7= 2^3 -2(2)^2-2+9
7 = 8-8-2+9
7=7

b) Find the equation of the tangent to C at P, giving your answer in the form y=mx+c, where m and c are constants.

found to be ;
dy/dx= 3x^2-4x - 1
12-8-1
m = 3
y-7=3(x-2)
y = 3x+1
The point Q also lies on C

Given that the tangent to C at Q is perpendicular to the tangent to C at P,

c) show that the x coordinates of Q is 1/3 (2 + square root 6)
??
So first find the perpendicular gradient.
Set
3. (Original post by shaunobrien1)
I've tried this question out and I've done a and b but can't do part c.

The curve C has equation y=x^3 - 2x^2 - x + 9 x>0

The point P has coordinates (2,7)

a) Show that P lies on C

worked out to be ;
7= 2^3 -2(2)^2-2+9
7 = 8-8-2+9
7=7

b) Find the equation of the tangent to C at P, giving your answer in the form y=mx+c, where m and c are constants.

found to be ;
dy/dx= 3x^2-4x - 1
12-8-1
m = 3
y-7=3(x-2)
y = 3x+1
The point Q also lies on C

Given that the tangent to C at Q is perpendicular to the tangent to C at P,

c) show that the x coordinates of Q is 1/3 (2 + square root 6)
??
For part C you already know the gradient of the tangent to C at P, it's 3 (assuming you did it correctly). So you want the gradient at point Q to be -1/3 since it's perpendicular to the tangent at P.
So you're literally just solving .

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