Find the equation of the tangent and the normal line to the curve y = (1+2x)^2 at point (1,9)?

Write the parabola equation in standard form. Y = (1 + 2x)² y = 4(x + 1/2)² (x + 1/2)² = (1/4)y Vertex: (-1/2, 0) Axis: x = -1/2 The projection of (2, 25) onto the axis is (-1/2, 25). The point on the axis the same distance from the vertex, on the opposite side is (-1/2, -25).

The tangent line includes points (2, 25) and (-1/2, -25). (y - 25)/(x - 2) = (-25 - 25)/(-1/2 - 2) (y - 25)/(x - 2) = 20 (y - 25) = 20(x - 2) y - 25 = 20x - 40 20x - y - 15 = 0.

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