Of course not! There's a whole x-y Cartesian plane out there to play on, so there's no reason they should be. They can be tilted and positioned any which way. Of course, when you start out learning about conic sections (including parabolas), you start with the simplest parabola, y = x^2, then graduate to (y-b) = (x-a)^2 and similar forms interchanging x and y. And when you use parabolas, to figure the trajectory of an object which was launched with a certain velocity, in a certain direction, and is subject to gravity, you use only untilted parabolas (well, parabolae, if you want to use the correct plural word).
Most algebra classes don't have you figure out what a tilted parabola equation would look like, but it isn't that tough to figure out. First, though, it helps to reconsider just what it is that you do when you graph a point on the curve of an equation (and you thought you knew!).
Let's say you're working with an ordinary function on ordinary x-y axes (it could be a parabola or anything else!). To graph a point at (3,5), let's say, you proceed +3 units distance along your first axis (the x-axis, which you might also call the 1x + 0y axis). You are at (3,0).This gives you +3 for the coordinate along your first axis.
Then, you turn at a right angle, and proceed +5 units along your second axis (the y-axis, which you might also call the 0x + 1y axis). That gives you +5 for the co-ordinate along your second axis. And there you are at your desired point, (3,5).
So now let's try this process for a graph in which the desired graphing axes (let's call them new x-axis, and new y-axis; usually textbooks will represent them as x' and y', which may confuse you if you know other definitions for y' of a function already) are set at 45 degrees clockwise from the normal x-y axes ("rotated clockwise" or "tilted right").
First, we proceed +3 units along a line dropped 45 degrees from the horizontal. Whoa, where are we now? You may have to solve a little Pythagorean equation to get (root(9/2),-root(9/2)). Now, compare this intermediate point to the intermediate point (3,0) two paragraphs above. In the first example, by proceeding along the x-axis only, we added value to x only, and added no value to y. So our first coordinate had the form of x (and nothing relating to y). But in this "tilted right" example, we added value to x, and subtracted value from y (where x and y are considered according to the old axes). So we can write an expression, (x-y), which represents what we did (i.e., where we moved) (where x and y stand for the old-axes coordinates). Since x is positive signed in the expression, if we added value to x (which we did), we added value to the expression (moved positive along the new "x"-axis); if we subtracted value from y (which we did, it's now -root(9/2)), because y appears with a negative sign in the expression, we also *added* value to the expression (- x - = +, two negatives multiply to make a positive). So, the expression (x-y) represents what we did by moving along the first new axis. (Well, actually, there's a multiplier constant also, but we'll just leave that out, it doesn't affect the form of the results we'll get.)
Now here's the important part. The new axis represents the new argument (first variable) of whatever function we were graphing. So anything that stands for the distance along this axis, stands for the value of our first variable. Which means, we can substitute the expression (x-y) in to stand for the first variable of the new, tilted graph. In fact any place we had an "x" in the old function equation, we'll put "(x-y)".
Next, we graph the new axis "y" value. From where we were at, we proceed +5 units along a line inclined 45 degrees above horizontal. This will add value to both x and y, root(25/2) for each.
By comparable consideration to what we did for the new-x-axis value, we also make an expression for the new-y-axis value: (x+y). And, we can then use this expression to stand for the second variable (which was "y" as it appeared in the old function). Any place we had a "y" in the old function equation, we'll put a (x+y) in the new equation.
The point (3,5) when graphed using our new axes, still stands for (3,5) as measured in our new axis system! But in our old axis system, it is the point (((root(9/2)+root(25/2)),(-root(9/2)+root(25/2))). This looks incredibly complicated, but we weren't after the values so much as the ideas behind the connection with variables and their axes.
What do we get, then, when we transform (rotate) the equation (and graph) for a parabola, from the old axes (regular x-y Cartesian coordinates) into the new axes?
In terms of the new axis, the equation is still (y-b) = (x-a)^2 !
But, substituting in what we just figured above, *in terms of the old axes*, things would look like:
((x+y)-b) = ((x-y)-a)^2
which expands to
x+y-b = ((x-y)^2 - 2a(x-y) +a^2) = x^2 - 2xy +y^2 -2ax +2ay +a^2
We can collect terms to better see what we have:
0 = x^2 + y^2 - 2xy - (2a+1)x + (2a-1)y + (a^2 + b)
In short, we now have a general equation for a conic section, with x^2, y^2, xy, x, y, and c terms.
This is what the equation of a rotated parabola looks like (for other amounts of rotation, as you might imagine, the coefficients of the new equation will be a different set of numbers than the set we just obtained above. But the form of the equation -- the types of terms it has -- will be like what we just obtained, except in the special cases where one of the original x and y variables maps back onto itself with sign flipped, or when the original x and y variables map back onto each other). And although it may look like a formidable job to ever simplify back to the original parabola equation statement, it is a job like many others you already know how to do -- complete the square of a quadratic expression, factor a quadratic expression by synthetic division, and so on (but maybe a little harder). Most likely, nowadays, you would let your graphing calculator display the graph, recognize it visually, and take it from there.
A big caution!! Hopefully, you'll be careful enough to examine the graph to make sure that you're seeing a parabola, rather than one branch of a hyperbola, or a tiny piece of an ellipse!
Enjoy your graphings!