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find the equation of the tangent line to \(y=16x-12x^2+2 \) at x=2

 Apr 9, 2020
 #1
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First:  find the y-value when x = 2:     y  =  16(2) - 12(2)2 + 2  =  -14   --->   (2, -14)

 

Second:  in order to find the slope of the line, find the first derivative of y:

      y'  =  16 - 2·12x     --->     y'  =  16 - 24x

At x = 2   --->   y'  =  16 - 24·2  =  -32

 

You know have both the point and the slope at that point; use the point-slope form:

     y - -14  =  -32(x - 2)

 Apr 9, 2020

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