Hello, usagi_killer!
Find the values of $\displaystyle p$ such that $\displaystyle y=px$ intersects $\displaystyle y=x^2 + 1$ twice.
Explain the geometric significance of this. The parabola and line intersect: .$\displaystyle x^2+1 \:=\:px \quad\Rightarrow\quad x^2  px + 1 \:=\:0$
. . Quadratic Formula: .$\displaystyle x \:=\:\frac{p \pm \sqrt{p^24}}{2}$
With two intersections, the discriminant must be positive:
. . $\displaystyle p^2  4 \:>\:0 \quad\Rightarrow\quad p^2 \:>\:4 \quad\Rightarrow\quad p \:>\:2 $
Hence: .$\displaystyle p \:<\:2\;\text{ or }\;p \:>\:2 $
Make a sketch.
We have an upopening parabola with vertex (0,1)
. . and a straight line through the origin. Code:
*  *

*  *
*  *
*  * o
1* o
 o
 o
 o
     +       
o 
o 

If the slope $\displaystyle p$ is +2 or 2, the line is tangent to the parabola.
. . There is one intersection.
If $\displaystyle p$ is greater than 2 and less than 2, the line misses the parabola.
. . There is no intersection.
If $\displaystyle p$ is less than 2 or greater than 2, there are two intersections.