# Thread: Finding equation of a tangent line given slope of tangent

1. ## 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?

2. 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.

3. You have the same answer as the textbook. Just multiply through by 8.

4. You need $y-f(x_0)=1(x-x_0)$ where $x_0$ is the solution to $8x^3=1$

5. Why don't I use the y-coordinate?