Rotate the parabola $y=x^2$ clockwise $45^\circ$. I used the rotation matrix to do this and I ended up with the equation: $$x^2+y^2+2xy+\sqrt{2}x-\sqrt{2}y=0$$
I tried to plot this but none of the graphing softwares that I use would allow it.
Is the above the correct equation for a parabola with vertex (0,0) and axis of symmetry $y=x$ ?
$$\left(
\begin{array}{cc}
\cos\theta&-\sin\theta\\
\sin\theta&\cos\theta\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
For a clockwise rotation of $\frac{\pi}{4}$, $\sin{-\frac{\pi}{4}}=\frac{-1}{\sqrt{2}}$ and $\cos{-\frac{\pi}{4}}=\frac{1}{\sqrt{2}}$
$$\left(
\begin{array}{cc}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
\end{array}
\right)\left(
\begin{array}{cc}
x\\
y\\
\end{array}
\right)=\left(
\begin{array}{cc}
X\\
Y\\
\end{array}
\right)$$
$$X=\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$Y=\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}$$
$$y=x^2$$
$$\left(\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)=\left(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}\right)^2$$
$$\frac{-x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{x^2}{2}+\frac{2xy}{2}+\frac{y^2}{2}$$
$$-\sqrt{2}x+\sqrt{2}y=x^2+2xy+y^2$$
$$x^2+2xy+y^2+\sqrt{2}x-\sqrt{2}y=0$$
Have I made a mistake somewhere? 

Before rotating (and plotting), you need to parametrize your parabola:
$$\begin{cases}
x(t) = t\\
y(t) = t^2
\end{cases},$$
where $t \in \mathbb{R}.$
You are rotating each point of the parabola, and hence:
$$\begin{bmatrix}X(t)\\Y(t)\end{bmatrix} = \frac{\sqrt{2}}{2}\begin{bmatrix}1 & -1\\1 & 1\end{bmatrix}\cdot\begin{bmatrix}x(t)\\y(t)\end{bmatrix} = \frac{\sqrt{2}}{2}\begin{bmatrix}x(t)-y(t)\\x(t)+y(t)\end{bmatrix}.$$
At the end you get that:
$$\begin{bmatrix}X(t)\\Y(t)\end{bmatrix} = \frac{\sqrt{2}}{2}\begin{bmatrix}t(1-t)\\t(1+t)\end{bmatrix}.$$
This can be plotted in Matlab using the following code:
t=linspace(-3,3,100);
X=(sqrt(2)/2)*(t.*(1-t));
Y=(sqrt(2)/2)*(t.*(1+t));
plot(X,Y)

This is what you get:

