Functions of several variables If $f(x,y) = x^2 + xy + y^2 - 3x + 4y - 5$. I know the domain is $\mathbb R^2$. How to determine the image of f is my issue.
 A: Have a look at my answer here: https://math.stackexchange.com/a/3619647/399263
You can always translate conics to cancel terms in $x,y$.
$f(x+a,y+b)=x^2+(-3+b+2a)x+xy+y^2+(2b+4+a)y+[\cdots]$
Solve $\begin{cases}2a+b-3=0\\2b+4+a=0\end{cases}\iff \begin{cases}a=\frac{10}3\\b=-\frac {11}3\end{cases}$
$$f(x+a,y+b)=\overbrace{(x^2+xy+y^2)}^{\ge 0}-\dfrac{52}3$$
In this case $x^2+xy+y^2\ge 0$ is always positive (discriminant $-3y^2<0$) with equality for $x=y=0$.
Thus $f$ has a minimum for $(x+a,y+b)=(0,0)\iff (x=-a,y=-b)$ of value $-\dfrac{52}3$.
Also $f$ is unbounded above since for instance $f(x,0)=x^2-3x-5$ is unbounded above.
A: Hint: This function is differentiable and has a unique critical point. It is helpful to determine whether this critical point is a saddle point, a maximum, or a minimum.
For instance, if the critical point is a local maximum and if that local maximum is also a global maximum, then the function attains a unique maximum but it unbounded below, which means that the range is of the form $(-\infty,c]$ for some $c \in \Bbb R$.

Alternatively, it would suffice to show that this function is convex.
A: Hint Consider the qadratic form
$$Q(x,y,z)= x^2 + xy + y^2 - 3xz + 4yz - 5z^2$$
Note that
$$f(x,y)=Q(x,y,1)$$
$Q$ corresponds to the symmetric matrix
$$A= \begin{bmatrix}
1 & \frac{1}{2} &\frac{-3}{2} \\
\frac{1}{2} & 1 & 2\\
\frac{-3}{2}&2 & -5 
\end{bmatrix}$$
Now, orthogonally diagonalize $A$, and do the corresponding change of variable $\begin{bmatrix} a \\ b \\ c \end{bmatrix} = P\begin{bmatrix} x \\ y \\ z\end{bmatrix} $ to transform $Q$ to the form
$$Q(a,b,c)=\alpha a^2 + \beta b^2+ \gamma c^2$$
Swicth $a,b,c$ back to $x,y,z$ and set $z=1$.
