Equation of common tangent of y = x2, y = –x2 + 4x – 4 i

The common tangent of the parabolas y = x^2 and y = -x^2 + 4x - 4 can be found by equating their derivatives and solving for x.

Step 1: Find the derivatives of the two functions.

The derivative of y = x^2 is 2x. The derivative of y = -x^2 + 4x - 4 is -2x + 4.

Step 2: Set the derivatives equal to each other and solve for x.

2x = -2x + 4 4x = 4 x = 1

Step 3: Substitute x = 1 into the original equations to find the y-coordinates of the points of tangency.

For y = x^2, y = (1)^2 = 1. For y = -x^2 + 4x - 4, y = -(1)^2 + 4(1) - 4 = -1.

Step 4: Use the point-slope form of a line to find the equations of the tangents.

The equation of the tangent to y = x^2 is y - 1 = 2(x - 1), or y = 2x - 1. The equation of the tangent to y = -x^2 + 4x - 4 is y + 1 = 2(x - 1), or y = 2x - 3.

Therefore, the common tangent of the two parabolas is y = 2x - 1.