Tangent line help If $f(x)= x^2+2$, find all the points on the graph of $f$ for which the tangent line passes through the origin $(0,0)$. So far I've used $2$ in place of $x$ and found the derivative is $y=4x-2$ but I don't know what to do from here. 
 A: Any line through the origin has the form $ y=mx$.
If you solve this simultaneously with the curve $y=x^2+2$ you get the quadratic equation $$x^2-mx+2=0$$
This must have double roots, so the discriminant is zero. This gives $m^2-8=0$ to give $m=\pm 2\sqrt{2}$ but also, as a double root, $$x=-\frac{b}{2a}=\frac m2=\pm \sqrt{2}$$
A: suppose the line $y = mx$ through the origin $(0,0)$ is tangent to $y = x^2 + 2.$  then the $x$-coordinates of the points of intersection are give by $$x^2 + 2 = mx\to x^2 - mx + 2 = 0. $$ the above quadratic equation should have a repeating root for $y = mx$ to be tangent. the condition is that the discriminant $$(-m)^2 - 4 \times 1 \times 2  = 0 \to m = \pm 2\sqrt 2. $$ 
A: The slope of the line that passes through $P(x,x^2+2)$ and $O(0,0)$ is equal to the derivative of $x^2+2$.
$$m = \frac{(x^2+2) - 0}{x-0} = 2x$$
$$x^2+2=2x^2$$
$$x^2=2\implies x=\pm\sqrt{2}\implies P=(\pm\sqrt{2}, 4)$$
A: If you want to use derivatives: the tangent line will have the form $y=\lambda x$ where $\lambda=f'(x_0)$ We know $f'(x_0)=2x_0$ (you can calculate this from the definition of the derivative) and $(x_0,y_0)$ is the intersection point(s). Both $y_0=f(x_0)=x_0^2+2$ and  $y_0=\lambda x_0=[f'(x_0)]x_0=2x_0^2$ should hold and so $x_0^2=2$ which gives the answer. 
