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# C. f(x)=8x 2 -4x+3 = 16x-4

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C. f(x)=8x2-4x+3 = 16x-4

use c. to find the tangent line at x=1. Where is there a horizontal tangent line?

May 15, 2018

### 1+0 Answers

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The slope of the tangent line at  x  = 1  is   16(1) - 4   = 12

And at x  = 1, then  y  =  8(1) ^2  -4(1)  + 3   =  8 - 4 + 3  = 7

So...the equation of the tangent line is

y  = 12 (x - 1)  + 7

y = 12x  - 12 +  7

y  =12x - 5

Here's the graph : https://www.desmos.com/calculator/daj2o31drm

The horizontal tangent line will be tangent to the vertex.......it will have a slope of  0

So...we have that the x coordinate of the vertex is

16x - 4  = 0

16x  = 4

x  = 1/4

And when x  = 1/4, y  =  8(1/4)^2 - 4(1/4) + 3  =  1/2 - 1 + 3  =  2.5  = 5/2

So.....there is a horizontal tangent line at  ( 1/4, 5/2 )   May 15, 2018