Please help

1.

Let f(9:) = V70 a: The slope of the tangent line to the graph of f(a:) at the point (6, 8) is: C] Hint: the slope is given by the derivative at a: = 6, ie. ( . f(6+h)f(6)) hm h>0 h Let f(:c) = 49:2 13:6 + 3 The slope of the tangent line to the graph of f(:1:) at the point (4, 15) is C] . The equation of the tangent line to the graph of at) at (4, 15) is y = ma: + b for and Hint: the slope is given by the derivative at a: = 4, ie. (H f(4+h) f(4)) m h>0 h Let me) = % 10 The slope of the tangent line to the graph of f(:I:) at the point (9, F) is: :] Hint: the slope is given by the derivative at a: = 9, ie. (H f(9+h) f(9)) m h>0 h Let 1 m = ) as + 2 Use the limit definition of the derivative to find ww = :] = :i W = :] = :1 To avoid calculating four separate limits, I suggest that you evaluate the derivative at the point when :5 = 0.. Once you have the derivative, you can just plug in those four values for "a" to get the answers. The volume, v, in cubic meters of a leaking tank of water after t seconds is modeled by the function 405 v t = ( ) t + 1 answer to two decimal places if necessary. m3 :1 3 for t Z 0. Find the rate at which the water is leaving the tank after 2 seconds. Round your Let f(a:) = 3:32 3s: + 4 The Slope-Intercept Form of the line tangent to f($) at (2, 10) is: y = C]