We determine all $p$ such that the graphs of the functions $y_1(x)=\frac{1}{4}x^2+px$ and $y_2(x)=2x-p^2$ intersect at least once.
The graphs intersect if the function values are the same. Hence, we start by solving the equation $y_1(x)=y_2(x)$.
$\frac{1}{4}x^2+px=2x-p^2 \Leftrightarrow \frac{1}{4}x^2+(p-2)x+p^2=0$.
Define the quadratic function $f(x)=\frac{1}{4}x^2+(p-2)x+p^2$. Note that the discriminant of $f(x)$ depends on $p$. The function $f(x)$ has one zero if the discriminant is zero, two zeros if the discriminant is positive and no zeros if the discriminant is negative. Hence, we have to find all $p$ such that $D \geq 0$. We find the zeros of $D$ as follows:
$$\begin{align}
D(p)= 0 &\Leftrightarrow (p-2)^2-4\cdot \frac{1}{4}\cdot p^2=0\\
&\Leftrightarrow -4p+4=0\\
&\Leftrightarrow p=1.
\end{align}$$
From the sign chart of $D(p)$ it easily follows that for $p \leq 1$ it holds that $D\geq 0$ and hence, the graphs intersect at least once.
