# Math Help - Finding horizontal tangent lines

1. ## Finding horizontal tangent lines

1. find each value of x at which $f(x) =x(2x-8)^3$ has a horizontal tangent line.

$y'=(2x-8)^3 + 6x(2x-8)^2$

set = to 0
$(2x-8)^3 + 6x(2x-8)^2=0$

factoring out
= $[(2x-8)^2][(2x-8)+(6x)]$

= $(2x-8)=0$
x=4
= $(6x)=0$
x=0
Key says the answer is x=1, and x=4.

2. Originally Posted by hazecraze
= $[(2x-8)^2][(2x-8)+(6x)]$

= $(2x-8)=0$
How did you arrive at this equality? How did you go from "2x - 8 = 0" to "2x = 0 and 6x = 0"?

$(2x\, -\, 8)^2 (8x\, -\, 8)$