# Find the intersection point of the graph x^2 -3 and the line y= 2x.

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At the points where the graphs y = x^2 - 3 and y = 2x intersect, the x and y coordinates are the same.

Here we substitute 2x for y in y = x^2 - 3

=> 2x = x^2 - 3

=> x^2 - 2x - 3 = 0

=> x^2 - 3x + x - 3 = 0

=> x(x - 3) + 1( x - 3) = 0

=> (x + 1)(x - 3) = 0

=> x = -1 and x = 3

y = 2x = -2 and y = 6

**The required points of intersection are (-1, -2) and ( 3, 6)**

Given the curve x^2-3 and the line y= 2x

We need to find the intersection points.

==> x^2 - 3 = 2x

==> x^2 - 2x - 3 = 0

Now we will factor.

==> (x-3)(x+1) = 0

==> x1= 3 => y1= 2*3 = 6

==> x2= -1 ==> y2= 2*-1 = -2

Then we have two intersection points for the curve and the line :

**The points of intersection are: ( 3, 6) and (-1, -2).**