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Unformatted text preview: 3 x ( x4) =3 x = x4 = x = x = 4 Now to find the points, you must find the corresponding yvalues for these xvalues by substituting back into x=0 x=4 Thus, the tangent line to the graph of f is horizontal at the points (0, 0) and (4, 32). f ( x ) = x 3 + 6 x 2 . f (0) = 3 + 6 ⋅ 2 f (0) = f (4) = 4 3 + 6 ⋅ 4 2 f (4) = 32 Graph for this example Write the equation of the tangent line at a given point on the graph Write the equation of the tangent line to the graph of at the point (3,9) First find the slope using m = . For this function . At the point (3,9), f x ( 29 = x 2 , ′ f x ( 29 = 2 x ( 29 ( 29 3 2 3 6 m f ′ === ( 29 f x ′ Now use m = 6 and the point (3,9) to write the equation for the tangent line ( 29 ( 29 y x y x y x y x =   = + = = 9 6 3 9 6 3 9 6 18 6 9 Thus the tangent line to the graph of at the point (3,9) is the line f x ( 29 = x 2 , y x = 6 9...
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This document was uploaded on 04/03/2011.
 Spring '09

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