At what point(s) on the curve x = 3t^2 + 4, y = t^3 − 6 does the tangent line have slope 1/2 ?

Answer 1

dy/dx = (dy/dt) / (dx/dt) = (3t²) / (6t) = (1/2)t

dy/dx = 1/2 when (1/2)t = 1/2. So t = 1

Therefore, x = 3(1)² + 4 = 7 and y = (1)³ - 6 = -5

(x, y) = (7, -5)