This is one of the most celebrated applications of completing the square. $$y=ax^2+bx+c \Longrightarrow y=a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)$$ Since a square (in your non-complex context) must be nonnegative, and will have a minimum when it is zero, the vertex of the parabola will be at $$(x,y)=(-b/2a,c-b^2/4a)$$ If this equation doesn't look like the stretched parabola $ax^2$ shifted by this vertex, what would?
Also, you can find the places it hits the $x$-axis by solving for $y=0$ using the quadratic formula or factoring. The $y$-intercept would be $c$, of course. Hopefully with these three pieces of information you could plot as much as you want.
By the way, this isn't really a matheducators.SX type answer, but I hope it leads you to thinking about the many problems typically assaulted with calculus that may not need it, proving that dogs don't need calculus.