# tangent line equations

• February 5th 2010, 02:46 PM
phalange
tangent line equations
Find an equation of the line that is tangent to the graph of f(x)=x^3 and parallel to the line: 12x-y+5=0
• February 5th 2010, 02:55 PM
e^(i*pi)
Quote:

Originally Posted by phalange
Find an equation of the line that is tangent to the graph of f(x)=x^3 and parallel to the line: 12x-y+5=0

Rearrange the second equation into y=mx+c to find the gradient

$y = 12x+5$ therefore the gradient is 12

If the tangent [ie: f'(x)] is parallel to this line it must be equal.

Find f'(x) and set it equal to 12 then get x.

Spoiler:
$f'(x) = 3x^2 = 12$
$x = \pm 2$

Put your value(s) of x into f(x) to get a co-ordinate.

Spoiler:
$f(-2)= -8 \: , \: f(2) = 8$

You can now use the equation of a straight line $y-y_1 = m(x-x_1)$

To my knowledge I am not sure if you can get an explicit solution