# Find the point(s) of intersection of the parabola with equation y = x2 - 5x + 4 and the line with equation y = 2x - 2

*print*Print*list*Cite

x2 - 5x + 4 and the line with equation y = 2x -2

Let the parabola be:

f(x) = x^2 - 5x + 4

and the line y = 2x -2

The point where f(x) and y intersects is when f(x) = y

Let us susbtitue:

f(x) = y

==> x^2 - 5x + 4 = 2x - 2

Combine like terms:

==> x^2 - 5x - 2x + 4 + 2 = 0

==> x^2 - 7x + 6 = 0

Now we will factor:

==> x^2 - 7x + 6 = ( x-6)(x-1)

Then:

x1 = 6 ==> y1 = 2(6) -2 = 10

x2= 1 ==> y2 = 2(1) -2 = 0

**Then the parabola f(x) and the line y intersects at two points:**

** ( 6, 10) and ( 1, 0)**

The common point that lies on the line and parabola in the same time is the intercepting point of the line and parabola.

So, the y coordinate of the point verify the equation of the line and the equatin of the parabola, in the same time.

2x-2 = x^2-5x+4

We'll factorize by 2 both sides:

2(x-1) = x^2-5x+4

But the roots of the quadrstic are x1 = 1 and x2 = 4.

2(x-1) =(x-1)(x-4)

We'll divide by (x-1):

2 = x - 4

We'll subtract 2 both sides and we'll use the symmetric property:

x - 6 = 0

x = 6

Now, we'll substitute in the equation of the line to determine y:

y = 2x-2

y = 2*6 - 2

y = 10

**The common point between the line and parabola is (6 , 10).**