After viewing the equation in desmos and shifting the value of k with a slider, I noticed that the parabola seems to hold the same shape, and to only translate while still touching the origin.
$$y = x^2 + kx - x$$
It might seem obvious that its shape indeed doesn't morph, but I cannot prove it. It seems to me that a possible proof would involve some sort of analysis to the formula of a parabola. The following formula is the general formula for all conic sections: (taken from Wikipedia)
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$
Which for our specific case would be:
$$x^2 + kx - x - y = 0$$
Note that the constant $D = k$
Another plausible approach would be the use of matrix transformations, but I am not proficient with the subject.
A translation on Cartesian plane is an isometry mapping $T:\Bbb R^2\rightarrow\Bbb R^2$ defined by $T(x,y)=(x+h_1,y+h_2)$ for some constants $h_1,h_2\in\Bbb R$. Then the image of the curve $f(x,y)=0$ under this translation is another curve given by the equation $f(T^{-1}(x,y))=f(x-h_1,y-h_2)=0$ and it has the same shape of the original curve. They are called congruent.
Let us translate the curve $y=x^2$ by the translation with $h_1=-\frac{k-1}{2}$, $h_2=-\frac{(k-1)^2}{4}$: $$y-h_2=(x-h_1)^2$$ $$y-\left(-\frac{(k-1)^2}{4}\right)=\left(x-\left(-\frac{k-1}{2}\right)\right)^2$$ $$y+\frac{(k-1)^2}{4}=\left(x+\frac{k-1}{2}\right)^2$$ $$y=x^2+(k-1)x$$ $$y=x^2+kx-x$$