The co-ordinate : (0;0) is the first solution, as pointed out by "Sciencesolve" as both these equations show that y=0 having no constant in either equation. In other words c (or q) is excluded from both equations:

`y = kx + c` and expanding the other equation `y= x^2 - 4x + c`

Determine another reference point on your parabola`` other than (0;0)

If y= -3 ` therefore 0 = x^2 -4x + 3`

```(x-1)(x-3)=0`

`x=1 x=3`

(1;-3) and (3;-3). You will use this information later.

The line cuts the curve and therefore at those points the equations equal each other. Thus:

`kx= x^2 - 4x`

`therefore k=(x^2-4x)/x`

divide by the denominator

`therefore k=x-4`

We have established that x=3 is on the parabola so substitute into

`k=(3) -4`

`therefore` k=-1

Now find y in the original straight line y=kx when k=-1 and x=3

`therefore y=(-1)(3)`

`therefore y=-3`

`therefore ` (3;-3) is a co-ordinate on the straight line

We now know that our straight line equation is y=-x and the point other than (0;0) that is on both is (3;-3), having established above that (3;-3) is also a co-ordinate on the parabola.

We are looking for the range which is represented along ther y-axis

`therefore` `0<=y` `<=3`