f(x) = 1/x2 = x-2
f(-4) = 1/16
f'(x) = -2/x3
mtan = f'(-4) = 1/32
y - 1/16 = 1/32(x + 4)
Tanner B.
asked 02/09/22the limit of the difference quotient giving us the slope of the tangent line
f(x) = 1/x2 = x-2
f(-4) = 1/16
f'(x) = -2/x3
mtan = f'(-4) = 1/32
y - 1/16 = 1/32(x + 4)
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