# Finding equation of a tangent line given slope of tangent

• Jan 12th 2011, 01:44 PM
Apathy
Finding equation of a tangent line given slope of tangent
This question should be easy but it's giving me some trouble. I am asked to find the equation of the tangent to $f(x)=2x^4$ and told that the slope of the tangent is 1.

It is worded exactly: "Find the equation of a the tangent line to the graph $f(x)=2x^4$ that has slope 1.

So I found the derivative of f(x) to be $f'(x)=8x^3$

I then substituted f'(x) with 1 and found that x=1/2.

I substituted 1/2 into the original equation for x and found that f(x)= 1/8.

I then used (y-1/8)=1(x-1/2) and found the equation of the tangent to be y=x-3/8

The answer in my book was 8x-8y-3=0

Where did I go wrong?
• Jan 12th 2011, 01:49 PM
Also sprach Zarathustra
Quote:

Originally Posted by Apathy
This question should be easy but it's giving me some trouble. I am asked to find the equation of the tangent to $f(x)=2x^4$ and told that the slope of the tangent is 1.

It is worded exactly: "Find the equation of a the tangent line to the graph $f(x)=2x^4$ that has slope 1.

So I found the derivative of f(x) to be $f'(x)=8x^3$

I then substituted f'(x) with 1 and found that x=1/2.

I substituted 1/2 into the original equation for x and found that f(x)= 1/8.

I then used (y-1/8)=1(x-1/2) and found the equation of the tangent to be y=x-3/8

The answer in my book was 8x-8y-3=0

Where did I go wrong?

Multiply the last equation by 8.
• Jan 12th 2011, 01:50 PM
Plato
You have the same answer as the textbook. Just multiply through by 8.
• Jan 12th 2011, 01:51 PM
pickslides
You need $y-f(x_0)=1(x-x_0)$ where $x_0$ is the solution to $8x^3=1$
• Jan 12th 2011, 01:56 PM
Apathy
Why don't I use the y-coordinate?
• Jan 12th 2011, 01:57 PM
Plato