I know how to reflect a coordinate over the $y$ and $x$ axis, but is there a rule I could use to help me find the reflected point over $x = -1$?
This is what I know already:
Over the $x$-axis: $(x, y) \to (x, –y) $
Over the $y$-axis: $(x, y) \to (–x, y)$
Over the line $y = x$: $(x, y) \to (y, x)$
Through the origin: $(x, y) \to (–x, –y) $
What I don't know is how to solve when reflecting over something like $x = -1$. Is there a rule that would help make solving this easier?
On
Rather than think about transformation rules symbolically, and trying to generalize them, try thinking about them visually. The line $x = -1$ is a vertical line one unit to the left of the $y$-axis. (You should follow along and draw things out on a sheet of graph paper or on your computer, in order to make them clear.) Therefore, if you have a point at $(3, -5)$, it is three units to the right of the $y$-axis, but four units to the right of your axis of symmetry.
When you reflect this point, you should end up at the same "height" ($y$-coordinate) of $-5$, but this time four units to the left of your axis of symmetry. Four units to the left of $x = -1$ is $x = -1-4 = -5$, so the point $(3, -5)$ reflects to $(-5, -5)$. We might write
$$ (3, -5) \to (-5, -5) $$
Similar reasoning shows that, for example,
$$ (-2, 4) \to (0, 4) $$ $$ (0, 0) \to (-2, 0) $$ $$ (13, 2) \to (-15, 2) $$
So in each case, the $y$-coordinate stays the same, but $3$ becomes $-5$, $-2$ becomes $0$, $0$ becomes $-2$, and $13$ becomes $-15$. If you're a perceptive sort, you might notice that the sum of each of these pairs of $x$-coordinates is $-2$, and therefore arrive at the transformation rule $x' = -2-x$, but if not, you can still reconstruct what's happening.
The point $x$ is how far to the right of your axis of symmetry? The axis of symmetry has an $x$-coordinate of $-1$, so your distance to the right is $x-(-1)$, or $x+1$. (This distance is negative if you are actually to the left of the axis.) Then we need to move to the left by that amount. We do that by subtracting $x+1$ from $-1$, to get $x' = -1-(x+1) = -1-x-1 = -2-x$.
And of course, $y' = y$. So, finally,
$$ (x, y) \to (x', y') = (-2-x, y) $$
On
To reflect over a vertical line, such as $x=a$, first translate so the line is shifted to the y-axis, then reflect over it, then translate back so the line is shifted to its original position.
$$(x,y)\mapsto (x-a,y) \mapsto (a-x,y) \mapsto (2a-x,y)$$
In this case to reflex over $x=-1$ we shift $x\mapsto x+1$, reflect $\mapsto -1-x$ and shift back $\mapsto -2-x$
$$(x,y)\mapsto (-2-x, y)$$
Similarly for reflecting over horizontal lines, such as $y=b$ involves $(x,y)\mapsto (x, 2b-y)$.
Reflecting over a diagonal line is only a bit more complicated, as it involves rotation of the frame of reference.
Reflecting a point over $x=-1$ just means you have to keep the ordinate and let $x'-(-1)=-1-x \implies x'=-2-x$