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# Urgent Help Needed With Past Paper Questions Watch

1. The point Q lies on C and is such that the gradient of the normal to C at Q is 1/7.

Find the x coordinate of Q.

The curve C has the equation of y = x^2 -9x +13

--------------------------------

The coordinates for y=f(x) are

(-2, 0)
( 2,-3)
( 6, 0)

How can I draw y = -2f(x)
2. (Original post by Tempa)
The point Q lies on C and is such that the gradient of the normal to C at Q is 1/7.

Find the x coordinate of Q.

The curve C has the equation of y = x^2 -9x +13
If the gradient of the normal is 1/7 then what is the gradient of the tangent at Q?

Differentiate the function, make it = to the gradient above and solve for x
3. For 1 I got the x co-ord to be 1 , is that the answer ? . If it is i'll try go through it.

For 2, because it's -2, its a stretch for each by -2 in the Y direction. So multiply the Y co-ord by -2 .

(-2, 0) stays -2 , 0
( 2,-3) goes 2 , 6
( 6, 0) stays 6, 0
4. (Original post by gdunne42)
If the gradient of the normal is 1/7 then what is the gradient of the tangent at Q?

Differentiate the function, make it = to the gradient above and solve for x
The gradient of the tangent is 3 as dy/dx is 2x - 9

And the tangent is at point p to Curve C, at ( 6, -5)
5. (Original post by Tempa)
The gradient of the tangent is 3 as dy/dx is 2x - 9

And the tangent is at point p to Curve C, at ( 6, -5)
Do you understand what a normal is?
The normal at Q is perpendicular to the tangent at Q
6. (Original post by gdunne42)
Do you understand what a normal is?
The normal at Q is perpendicular to the tangent at Q
No sir I do not know.
7. (Original post by Tempa)
No sir I do not know.
You should know from coordinate geometry that if 2 lines are perpendicular then the product of their gradient = -1
or expressed another wayif m1 and m2 are the gradients of the 2 lines then m1 x m2 = -1

So if the gradient of the normal at Q is 1/7 the gradient of the tangent must be -7 (1/7 x -7 = -1)

Then 2x - 9 = -7

Does this make any sense, because if not then it's seems premature to be trying exam questions.

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Updated: December 20, 2010
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