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?
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 )