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# Points of intersection with 3 curves watch

1. Show that the curves y= 2x^2 +5x y= x^2 +4x +12 y=3x^2 +4x -6

have one point in common and find it's coordinates.

I set equation 2= to equation 3 to get: x^2 +4x +12= 3x^2 +4x -6

I then rearranged it to get 2x^2 -18 and set this equal to equation 1 to get: 2x^2 -18= 2x^2 +5x

I then rearranged this to get x^2 -9 = 2x^2 +5x, therfore x^2 +5x + 9 =0

But i cant seem to factorise this and the quadratic formula doesnt seem to work as there is no negative value for C. the answer is apparntly (3,33). Any help?
2. Let's say that curve 1 is , curve 2 is and curve 3 is . You're trying to find the point at which .

When you set eq 2 equal to eq 3, you had . You then rearranged to get ; and then for some reason you set this equal to eq 1, giving -- this clearly isn't right.

What you need to do is solve . You'll get two possible values of x. You can then substitute each of these back into or (it doesn't matter which) and you're left with two coordinates.

If either of these lies coordinates satisfies , then it is a common point of intersection for all three curves.
3. nuodai's method is the only way I can think of solving it
4. (Original post by nuodai)
Let's say that curve 1 is , curve 2 is and curve 3 is . You're trying to find the point at which .

When you set eq 2 equal to eq 3, you had . You then rearranged to get ; and then for some reason you set this equal to eq 1, giving -- this clearly isn't right.

What you need to do is solve . You'll get two possible values of x. You can then substitute each of these back into or (it doesn't matter which) and you're left with two coordinates.

If either of these lies coordinates satisfies , then it is a common point of intersection for all three curves.
Ah I see . So I solved g(x)=h(x) to get x = -3 or x = 3 so I then substitue x = 3 into f(x) to get 33 (I used f(x) as 2x^2 +5x in my working). I thne check this equals the same for g(x) also?
5. (Original post by nuodai)
Let's say that curve 1 is , curve 2 is and curve 3 is . You're trying to find the point at which .

When you set eq 2 equal to eq 3, you had . You then rearranged to get ; and then for some reason you set this equal to eq 1, giving -- this clearly isn't right.

What you need to do is solve . You'll get two possible values of x. You can then substitute each of these back into or (it doesn't matter which) and you're left with two coordinates.

If either of these lies coordinates satisfies , then it is a common point of intersection for all three curves.

Thanks btw eally well explained i get it now .
6. (Original post by hazbaz)
Ah I see . So I solved g(x)=h(x) to get x = -3 or x = 3 so I then substitue x = 3 into f(x) to get 33 (I used f(x) as 2x^2 +5x in my working). I thne check this equals the same for g(x) also?
For both of your roots (x=-3,3) evaluate g(x) first so you know what your 'target' is. If, then, f(3)=g(3) or f(-3)=g(-3) then whichever it is is your point of intersection.
7. You must check both points to confirm that the curves have one and only one point in common.
8. (Original post by ghostwalker)
You must check both points to confirm that the curves have one and only one point in common.
Yep Thanks.

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Updated: September 30, 2010
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