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I was working on this practice problem and can't get the right answer. It is:

If f(x)=(x^(2)+4)(x^(3)-2), write an equation for the tangent line to f(x) at x=1

Thank you for your assistance!

y = (x²+4)(x³-2) = x⁵+4x³-2x²-8

dy/dx = 5x⁴+12x²-4x

slope at (1, -5) = 5·1⁴+12·1²-4·1 = 13

point-slope equation for line of slope = 13 that passes through (1, -5):

y+5 = 13(x-1)

in slope-intercept form:

y = 13x-18

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