# Tangent lines to a f(x)

1. Sep 28, 2006

### teken894

Seems like a simple problem, but I have no idea how to accomplish this!

f(x) = x^2 +4

I have to find a tangent line to the graph that passes through the point (0,0).

I know the derivative is 2x, but I don't know how to mathematically figure out the which (x,f(x)) the tangent line would pass through..!
Any help would be appreciated

2. Sep 28, 2006

If the derivative is $$2x$$, then the slope at $$(0,0)$$ will be $$0$$. So its just a horizontal line through $$(0,0)$$

3. Sep 28, 2006

### teken894

I know that.

You seee (0,0) is not a point on the graph. (0,0) is a given point (outside the graph) which I need to find a line tangent to the graph with.

4. Sep 28, 2006

you have to use Newtons method $$x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})}$$

5. Sep 28, 2006

### Rozenwyn

You could do it in a simpler way.
$$f(x) = x^2 + 4$$
$$f'(x) = 2x$$

Let $$g(x) = f'(x).x$$, i.e. write the equation of the tangent line in terms of x and $$f'(x)$$.

So you need a point where $$f(x) = g(x)$$
So: $$x^2 + 4 = (2x)x$$

Last edited: Sep 28, 2006
6. Sep 29, 2006

### HallsofIvy

Staff Emeritus
The problem did not say "tangent at (0,0)".

First, distinguish between the variable, x, and the x-value of the point of tangency. Let's call the x value at the point of tangency x0. Then the derivative at that point is 2x0 and so the tangent line has equation y= 2x0x. In order to be a tangent, that line must touch the parabola at x0: at x= x0 we must have y= 2x02= x02+ 4. Because of the symmetry, there are two lines through (0,0) tangent to y= x2+ 4