calculus
posted by vicki .
find an equation of the tangent line to the curve at y=2x^3+5x, (1, 3)

dy/dx = 6x + 5
so at (1,3), dy/dx = 6+5 = 1
slope = 1, point on line is (1,3)
3 = (1)(1) + b in y = mx+b
2 = b
equation of tangent:
y = x + 2
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