I'll use the parametrization
$$\begin{align*}x&=2at\\y&=at^2\end{align*}$$
where $a$ is the focal length (the distance from vertex to focus).
Using the formula for a parallel curve of $(f(t)\quad g(t))^T$ at a distance $c$:
$$\begin{pmatrix}f(t)\\g(t)\end{pmatrix}+\frac{c}{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}\begin{pmatrix}g^\prime(t)\\-f^\prime(t)\end{pmatrix}$$
we find the parametric equations
$$\begin{align*}x&=2at+\frac{ct}{\sqrt{1+t^2}}\\y&=at^2-\frac{c}{\sqrt{1+t^2}}\end{align*}$$
The corresponding Cartesian equation is rather complicated:
$$\begin{align*}x^2 \left(-8 a^3 y+x^2 \left(a^2-10 a y-3 c^2+y^2\right)-20 a^2 c^2+32 a^2
y^2+2 a c^2 y-8 a y^3+3 c^4-2 c^2 y^2+x^4\right)&=\\(c-y) (c+y) \left(4 a(a-y)+c^2\right)^2\end{align*}$$
so you're better off sticking to a parametrization.
Here's a plot of a bunch of parabola parallels:
What your teacher was referring to is the fact that:
$$x=\dfrac{a t^2+bt+c}{gt^2+ht+k}, y=\dfrac{d t^2+et+f}{gt^2+ht+k}, \ \ \ 0 \le t \le 1 \tag{1}$$
(with real coefficients $a,b,c,d,e,f,g,h,k$)
is in general a conic curve under the form of a rational quadratic Bezier curve.
This property comes from the vast domain of projective geometry.
The classification is easy: either the denominator
can be zero twice (which happens when its discriminant $\Delta=h^2-4gk > 0$) it means that there are two points at infinity: you have a hyperbola (a hyperbola has 4 points at infinity, but in projective geometry opposite points are considered as identical).
can be zero once ($\Delta=0$): parabola.
can never be zero ($\Delta<0$): ellipse.
In order to retrieve the driving points $A,B,C$, just decompose the numerators onto the basis $(1-t)^2, 2t(1-t), t^2$, which means finding $x_A, x_B,...$ such that:
$$\begin{cases}a t^2+bt+c&=&(1-t)^2 x_A + 2t(1-t)x_B+t^2x_C\\d t^2+et+f&=&(1-t)^2 y_A + 2t(1-t)y_B+t^2y_C\end{cases}$$
Remark: Any bijective change of parametrization $T=f(t)$ in (1) such that
$$f(0)=0 \ \ \text{and} \ \ f(1)=1\tag{2}$$
is possible. Taking a "Möbius" transformation
$$T=\frac{t}{rt+(1-r)}$$
complying with (2) preserves the form of equations (1) and allows, for an adequate choice of parameter $r$, to obtain (slightly) simpler expressions.
Recall: Equations (1) can be written under the form:
$$\underbrace{\begin{bmatrix}
X\\
Y\\
Z\end{bmatrix}}_{C'}=\underbrace{\begin{bmatrix}
a & b & c\\
d & e & f\\
g & h & k
\end{bmatrix}}_H \underbrace{\begin{bmatrix}
t^2\\
t\\
1 \end{bmatrix}}_C \tag{2}$$
followed by :
$$x=X/Z \ \ \text{and} \ \ y=Y/Z \tag{3}$$
Equation (2) expresses that one takes the image of "standard" parabola with parametric description given by vector $C$ (with so-called "homogeneous coordinates" : please note the third coordinate equal to $1$) to which an "homography matrix" $H$ is applied. Equations (3) are the classical homogenizing ratios by the third coordinate. See Fig. below.
Knowing that the image by a homography (either under the form (1) or (2)+(3)) of a conic curve is a conic curve.
See as well here.
Best Answer
Already so many answers, but I haven't seen my favorite one posted, so here's another.
The vertex occurs on the vertical line of symmetry, which is not affected by shifting up or down. So subtract $c$ to obtain the parabola $y=ax^2+bx$ having the same axis of symmetry. Factoring $y=x(ax+b)$, we see that the $x$-intercepts of this parabola occur at $x=0$ and $x=-\frac{b}{a}$, and hence the axis of symmetry lies halfway between, at $x=-\frac{b}{2a}$.