show that these curves

y=2x^2+5x
y=x^2+4x+12
y=3x^2+4x-6
have one point in common and find it coordinates??

is this done by simultaneous exuasion if so can you show me how because ive try and cant get it to work??

You can write two independent equations for x. Both must be satisfied for there to be a common solution. For example, from the first two of your equations:

2x^2 + 5x = x^2 + 4x +12
which can be rewritten
x^2 +x -12 = (x+4)(x-3) = 0
That has solutions of x=-4 and x=3.

You can also write:
x^2 + 4x + 12 = 3x^2 +4x -6
which is equivalent to
2x^2 = 18,
the solutions of which are
x = 3 or -3

Apparently there is a point at x = 3 that satisifies all of your equations. The y value at that point is 33. You can verify that with any of your first three equations.

So the answer is (3,33)

To find the common point(s) between the given curves, you can indeed use simultaneous equations. Here's how you can do it:

Step 1: Set up the equations for the given curves:
y = 2x^2 + 5x ...(Equation 1)
y = x^2 + 4x + 12 ...(Equation 2)
y = 3x^2 + 4x - 6 ...(Equation 3)

Step 2: Equate pairs of equations to form a system of equations. You can choose any two equations to start with. Let's choose Equation 1 and Equation 2:
2x^2 + 5x = x^2 + 4x + 12
Simplifying, we get:
x^2 + x - 12 = 0

Step 3: Solve the resulting quadratic equation to find the possible values for x. You can use factoring, completing the square, or the quadratic formula. In this case, the equation factors nicely:
(x - 3)(x + 4) = 0
So, either x - 3 = 0 --> x = 3
or x + 4 = 0 --> x = -4

Step 4: Substitute the values of x into any of the original equations to find the corresponding y-coordinates.

For x = 3:
Using Equation 1: y = 2(3)^2 + 5(3) = 2(9) + 15 = 18 + 15 = 33
So, one point in common is (3, 33).

For x = -4:
Using Equation 1: y = 2(-4)^2 + 5(-4) = 2(16) - 20 = 32 - 20 = 12
So, another point in common is (-4, 12).

Therefore, the two curves intersect at two points: (3, 33) and (-4, 12).