tangent line equations

**phalange** · February 5th 2010, 02:46 PM

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

**e^(i*pi)** · February 5th 2010, 02:55 PM

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:

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

Spoiler:

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

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