How Do You Find a Tangent Line to f(x) = x^2 + 4 That Passes Through (0,0)?

[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

 

[courtrigrad]

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

 

[teken894]

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

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.

http://66.93.135.68/tmp_pic.jpg

 

[courtrigrad]

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

 

[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

 

[HallsofIvy]

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

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