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...

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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`

You need to understand the request of the problem, hence, if the line `y=kx ` needs to cut the curve `y = x(x-4)` in two real distinct points, then the system of equations `{(y=kx),(y = x(x-4)):}` needs to have two real distinct solutions.

You should substitute `kx` for `y` in the equation of curve such that:

`kx = x(x-4)`

You need to open the brackets such that:

`kx = x^2-4x`

You need to move all terms to one side such that:

`x^2 - 4x - kx = 0`

You need to factor out x such that:

`x(x - 4 - k) = 0`

Hence, evaluating the solutions to the given equation yields:

`x=0 => y = 0`

`x = 4+k => y = k(4+k)`

**Hence, evaluating the solutions to the given system of equation yields `(0,0)` and `(4+k, k(4+k)), ` thus k may have any real value.**