Is an equation only a parabola if it is quadratic? Could the graph of $y = x^{1.65}$ be described as parabolic in shape? If given the equation $y = x^{1.65}$, could it be described as parabolic in shape, or does the equation have to have $x^2$ as its highest degree term?
 A: Not everything u-shaped is a parabola. A parabola has a very specific shape with particular  special properties.
For example, for each parabola there is a point called the "focus" and a line called the "directrix" and each point on the parabola is the same distance from the focus as it is from the directrix. This is similar to the way a circle has a center and every point of the circle is an equal distance from the center.
A shape might look more or less circular, but if it doesn't have a center that is the same distance from each of its points, it isn't a circle.
Curves such as $y=x^{1.65}$, don't have a focus and directrix that behave the way a parabola's do. They might look parabolic, but they are not parabolas.
Note also that an equation may have a parabolic graph even if it isn't obvious.  The answer below of
Vítězslav Štembera says that the equation must have one of two specific forms, but that isn't correct.  Any equation of the form $$Ax^2+2Bxy+Cy^2 +2Dx+2Ey+F=0$$ will have a parabolic graph, if it is non-degenerate and if $AC-B^2 = 0$.  For example, if one takes the parabola $y=x^2$ and rotates it by $45°$ the curve is still a parabola, with equation
$$x^2-2xy+y^2-x\sqrt 2-y\sqrt 2 = 0.$$
(See Rotate the parabola $y=x^2$ clockwise $45^\circ$. for details.)
A: A parabola is a function of the form $(y-y_0)^2=2p(x-x_0)$ or $(x-x_0)^2=2p(y-y_0)$ i.e. the quadratic term must be always present. $y=x^{1.65}$ is not a parabola.
