calculus
posted by Hannah .
Let f(x)=6x+16x^2 . Then the equation of the tangent line to the graph of f(x) at the point (2,16) is given by y=mx+b for
m= and b=

y = 16x^2 + 6x
y' = 32x + 6 for any x
The only problem I see is that y(2) = 76, so the point (2,16) is not on the graph.
So, now we have a line through (2,76) with slope y'(2) = 70
y76 = 70(x2) = 70x  140
y = 70x  64
Fix things up, and redo the steps above
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