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C. f(x)=8x^{2}-4x+3 = 16x-4

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

Guest May 15, 2018

#1**+1 **

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 )

CPhill May 15, 2018