How do you find the coordinates that the curves y= 2x^2 +5x, y=x^2+4x+12 and y=3x^2+4x-6 have in common?
The answer should be (3,33)
I know that you have to solve a simultaneous equations. What I've done so far:
y=2x^2+5x
y=x^2+4x+12 --> 2x^2+5x=x^2+4x+12
=x^2+x-12=0
=(x+6)(x-5)
= x=-6 or x=5
You want factors of 4 and -3 since they multiply to equal 12 and add to be the 1 in front of the x.
you should have x = 3 and x = -4
When I worked with the first and third, I got (x-3)(x+2)
x = 3 x = -2
so the x=3 showed up twice. Substitute it in each of the 3 equations to see if you get y=33 in each case.
In general, of course, three curves need not intersect in a single point. But, as you say,
2x^2+5x = x^2+4x+12: x = -4 or 3
2x^2+5x = 3x^2+4x-6: x = -2 or 3
x^2+4x+12 = 3x^2+4x-6: x = -3 or 3
So, it looks like (3,33) works fine
You messed up in your solution. 6*5 = 30, not 12!
http://www.wolframalpha.com/input/?i=plot+y%3D2x%5E2%2B5x,+y%3Dx%5E2%2B4x%2B12,+y+%3D+3x%5E2%2B4x-6,+-5+%3C%3D+x+%3C%3D+4
To find the coordinates where the three curves intersect, you need to solve the system of equations formed by equating each pair of curves.
1. Start by equating the first two curves:
y = 2x^2 + 5x ---(equation 1)
y = x^2 + 4x + 12 ---(equation 2)
2. Set the two equations equal to each other and solve for x:
2x^2 + 5x = x^2 + 4x + 12
2x^2 - x^2 + 5x - 4x - 12 = 0
x^2 + x - 12 = 0
3. Factor the quadratic equation:
(x + 4)(x - 3) = 0
This gives you two possible values for x: x = -4 or x = 3
4. Now substitute these values back into either of the original equations to get the corresponding y-values:
For x = -4:
y = 2(-4)^2 + 5(-4) = 32 - 20 = 12
For x = 3:
y = 2(3)^2 + 5(3) = 18 + 15 = 33
5. Hence, the coordinates that the three curves have in common are (-4, 12) and (3, 33).