Find the range of values of k for which the line y=kx will cut the curve y=x(x-4) at two real distinct points.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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` 

Approved by eNotes Editorial
An illustration of the letter 'A' in a speech bubbles

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.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial