How does the middle term of a quadratic $ax^2 + bx + c$ influence the graph of $y = x^2$?

Yes, it will effect both a horizontal and vertical translation, and you can see how much by completing the square. For example, $$x^2+3x=\left(x+\frac32\right)^2-\frac94$$

Compare that to your graph of $y=x^2+3x$. Of course, if the coefficient of the quadratic term is not $1$ things get a little more complicated, but you can always see what the graph the graph will look like by completing the square.

Look at $2$ Cartesian coordinate systems $X,Y$ and $X',Y'$.

Origin of $X',Y$' is located at $(x_0,y_0)$, $X'$-axis parallel $X$-axis , $Y'$-axis parallel $Y$-axis(a translation),i.e.

$x= x_0+x'$; $y= y_0+ y'$.

Set up your normal parabola in the $X',Y'$ coordinate system.

$y'=ax'^2$, vertex at $(0',0')$.

Revert to original $x,y$ coordinates .

$y-y_0= a(x-x_0)^2$ ;

$y=ax^2 -2(ax_0)x +ax_0^2$.

Compare with $y =ax^2+bc +c$:

$b=-2ax_0$.

Can you interpret?