# Math Help - Tangent Line

1. ## Tangent Line

if anybody could help with this that would be much appreciated:

Find the point on the graph f(x)=4/3x^3 where the tangent line is perpendicular to the line y=-16x+3

-Ryan

2. Originally Posted by compuryan
if anybody could help with this that would be much appreciated:

Find the point on the graph f(x)=4/3x^3 where the tangent line is perpendicular to the line y=-16x+3

Thus, $y=-16x+3$ means the perpendicular line has slope $1/16$.
The slope of any given point on the graph $f(x)=\frac{4}{3}x^3$ is its derivative which is,
$f'(x)=4x^2$.
We need that to be $1/16$ as explained in the first paragraph thus, $1/16=4x^2$. Solving we have that $x=\pm\frac{1}{8}$. These are the two x-values on the graph where the tangent is perpendicular to this line.