equation of tangent lines

[xxjeepdude44xx]

I've been banging my head over this one so if you guys could help me out I would greatly appreciate it!

Find the equation of 2 tangent lines to the graph f(x) = X² + 1 that pass through (0,0)

He's what i've tried so far:

F'(x) = 2x

(y-y₁)=m(x-x₁)
(y-0) = 2x(x-0)
x²+1=2x²
1=x²
x=sqrt(1)
=> f(x) = x²+1 = 1+1 = 2?

 

[DrPhil]

I've been banging my head over this one so if you guys could help me out I would greatly appreciate it!

Find the equation of 2 tangent lines to the graph f(x) = X² + 1 that pass through (0,0)

He's what i've tried so far:

F'(x) = 2x

(y-y₁)=m(x-x₁)
(y-0) = 2x(x-0)
x²+1=2x²
1=x²
x=sqrt(1)
=> f(x) = x²+1 = 1+1 = 2?

Your symbols are not clearly defined. If \(\displaystyle x\) is the independent variable, then you can't use the same name for a specific point on the graph. You may have done everything right, but I can't read it! Please be more precise in your statements.

How about

Let \(\displaystyle (x_1,\ y_1) = (x_1,\ x_1^2+1) =\) point of tangency for positive \(\displaystyle x\)

Then by symmetry, the other point of tangency is \(\displaystyle (-x_1,\ x_1^2+1)\)

Evaluate the derivative at \(\displaystyle x=x_1\). Then \(\displaystyle m = f'(x_1) = 2 x_1\)

But \(\displaystyle m\) is also the slope from \(\displaystyle (0,\ 0)\) to \(\displaystyle (x_1,\ x_1^2+1)\)

Equate the two ways to find \(\displaystyle m\), and solve for \(\displaystyle x_1\)