calculus
posted by Patty
Find the equation, in slope intercept form , of the line tangent to the graph of the square root of (x^3+7x+1), when x=3

Damon
when x = 3
y = (3^3 + 21 + 1)^.5 = sqrt(49) = 7 (or 7 but we will pretend we did not know that)
dy/dx = (1/2)(x^3+7x+1)^(1/2) (3 x^2)
at x =3 this is
(1/2)(1/7)(27) = 27/14 = slope of line
y = (27/14) x + b
7 = 27(3)/14 + b
solve for b
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