# Sketching Curves - C1 - Chapter 4

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Hello, can someone please help me out here...

How do you work out the equation of a curve using its points of intersection

How do you work out the equation of a curve using its points of intersection

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#2

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Hello, can someone please help me out here...

How do you work out the equation of a curve using its points of intersection

**Chittesh14**)Hello, can someone please help me out here...

How do you work out the equation of a curve using its points of intersection

Please post a question that you're stuck with.

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#4

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(x-A)(x-B)=0

**ubisoft**)(x-A)(x-B)=0

What would A and B be in that equation. They both have two coordinates, x and y coordinates. What is A in the bracket and what is B in the other bracket?

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#7

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I kind of understand that a bit.

What would A and B be in that equation. They both have two coordinates, x and y coordinates. What is A in the bracket and what is B in the other bracket?

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**Chittesh14**)I kind of understand that a bit.

What would A and B be in that equation. They both have two coordinates, x and y coordinates. What is A in the bracket and what is B in the other bracket?

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(Original post by

(x-A)(x-B)=0

**ubisoft**)(x-A)(x-B)=0

I understand that, but when I expand it, I get a different answer to the actual answer.

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The x coordinate of the intercept. The y intercept is 0 for both as it's the x intercept.

**ubisoft**)The x coordinate of the intercept. The y intercept is 0 for both as it's the x intercept.

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**ubisoft**)

The x coordinate of the intercept. The y intercept is 0 for both as it's the x intercept.

The answer in the book is x^2 + 2x - 5

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#11

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Maybe, the answer in the book is incorrect.

The answer in the book is x^2 + 2x + 5

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**Chittesh14**)Maybe, the answer in the book is incorrect.

The answer in the book is x^2 + 2x + 5

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#13

**Chittesh14**)

Maybe, the answer in the book is incorrect.

The answer in the book is x^2 + 2x + 5

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#14

use A first

substitute the x and the y coordinate of A into the quadratic and you will get a linear equation in p and q.

then you substitute the x and the y coordinate of B into the quadratic and you will get a second linear equation in p and q.

you solve simultaneously to get p and q

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(Original post by

if you have the coordinates of A and B

use A first

substitute the x and the y coordinate of A into the quadratic and you will get a linear equation in p and q.

then you substitute the x and the y coordinate of B into the quadratic and you will get a second linear equation in p and q.

you solve simultaneously to get p and q

**TeeEm**)if you have the coordinates of A and B

use A first

substitute the x and the y coordinate of A into the quadratic and you will get a linear equation in p and q.

then you substitute the x and the y coordinate of B into the quadratic and you will get a second linear equation in p and q.

you solve simultaneously to get p and q

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**TeeEm**)

if you have the coordinates of A and B

use A first

substitute the x and the y coordinate of A into the quadratic and you will get a linear equation in p and q.

then you substitute the x and the y coordinate of B into the quadratic and you will get a second linear equation in p and q.

you solve simultaneously to get p and q

I'm kind of confused here.

I've got two equations when I substituted the values from the A and the B coordinates.

A: -11 = px + q

B: -1 = px + q

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#18

(Original post by

I'm kind of confused here.

I've got two equations when I substituted the values from the A and the B coordinates.

A: -11 = px + q

B: -1 = px + q

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**Chittesh14**)I'm kind of confused here.

I've got two equations when I substituted the values from the A and the B coordinates.

A: -11 = px + q

B: -1 = px + q

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(Original post by

there should be no x, x was substituted with the x from each of the 2 coordinates from A and B

**TeeEm**)there should be no x, x was substituted with the x from each of the 2 coordinates from A and B

Q = -5 and p = 2

X^2 + 2x - 5

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