Find the point of intersection of the tangents to the curve y = x^2 at the points (-1/2, 1/4) and (1, 1).

To find the intersection of tangents to y =
x^2.

The equation of tangent to a curve at (x1,y1) is given
by:

y-y1 = (dy/dx)(x-x1).

So
dy/dx  = (x^2)' = 2x.

At x= (-1/2), dy/dx = 2(-1/2) =
 -1.

So the tangent at (-1/2, 1/4) is given
by:

 y-1/4 = (-1)(x+1/2).
Or

y-1/4 = -x-1/2.

y =
-x-1/2+1/4 = -x-1/4.

y =
-x-1/4...................(1).

Similarly at x = 1, dy/dx =
2x = 2*1 = 2. So the tangent at (1,1) is given by:

y-1 =
2(x-1)

=>y = 2x-2+1 = 2x-1.
Or

=>y =
2x-1...........(2).

y =
-x-1/4.........(1)

From (1) and (2) we
get:

2x-1 =
-x-1/4

=>2x+x = -1/4 +1 =
3/4.

=>3x=
3/4.

=> x = (3/4)/3 =
1/4.

Put x= 1/4 in eq (1) and we get  y = -x-1/4 = -1/4-1/4
= -1/2.

Therefore the point of intersection of tangents is
at (1/4, -1/2).