# Tangent Line

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• March 28th 2006, 01:29 PM
compuryan
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

Thanks in advance

-Ryan
• March 28th 2006, 06:24 PM
ThePerfectHacker
Quote:

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

Thanks in advance

-Ryan

Remember two lines are perpendicular to one another if and only if their slopes are negative reciprocals of each other.

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.