
# Classifying Critical Points


In order to develop a general method for classifying the behavior of a function of two variables at its critical points, we need to begin by classifying the behavior of quadratic polynomial functions of two variables at their critical points.

To see why this will help us, consider that the quadratic approximation of a function of two variables (its 2nd-degree Taylor polynomial) shares the same first and second partials as the function it approximates at the chosen point of tangency (or center point). Since sharing the same second partials means the two surfaces will share the same concavity (or curvature) at the critical point, this causes these quadratic approximation surfaces to share the same behavior as the function $$z = f(x, y)$$ that they approximate at the point of tangency. In other words, if the original function has a relative maximum at this point, so will the quadratic approximation. If the original function has a relative minimum at this point, so will the quadratic approximation, and if the original function has a saddle point at this point, so will the quadratic approximation.

Now there are really three basic behaviors of a quadratic polynomial in two variables at a point where it has a critical point. It will fit one of the following three forms, often being a transformation of one of the following functions.

1. A sum of two squared terms, like $$z = x^2 + y^2$$, producing a paraboloid that opens up and has a relative (absolute) minimum at its vertex. See the plot on the left side of Figure $$\PageIndex{4}$$.
2. The negative of a sum of two squared terms, like $$z = -\left(x^2 + y^2\right)$$, producing a paraboloid that opens down and has a relative (absolute) maximum at its vertex. See the plot on the right side of Figure $$\PageIndex{4}$$.
3. The difference of two squared terms, like $$z = f(x, y) = x^2 - y^2$$ or $$z = f(x, y) = y^2 - x^2$$, producing a saddle with a saddle point at its critical point. See Figure $$\PageIndex{3}$$.

Figure $$\PageIndex{4}$$: $$z = x^2 + y^2$$ has an absolute minimum of $$0$$ at $$(0,0)$$, while $$z = -(x^2 + y^2)$$ has an absolute maximum of $$0$$ at $$(0,0)$$,

Example $$\PageIndex{1}$$: Classifying the critical points of a function

Use completing the square to identify local extrema or saddle points of the following quadratic polynomial functions:

1. $$f(x,y) = x^2 - 6x + y^2 + 10y + 20$$
2. $$f(x,y) = 12 - 3x^2 - 6x - y^2 + 12y$$
3. $$f(x,y) = x^2 + 8x - 2y^2 + 16y$$
4. $$f(x,y) = x^2 + 6xy + y^2$$

Solution:

a. To determine the critical points of this function, we start by setting the partials of $$f$$ equal to $$0$$. \begin{align*} \text{Set}\quad f_x(x,y) &= 2x -6 = 0 & \implies x &= 3 \\ \text{and}\quad f_y(x,y) &= 2y + 10 = 0 & \implies y &= -5 \end{align*}We obtain a single critical point with coordinates $$(3, -5)$$. Next we need to determine the behavior of the function $$f$$ at this point.

Completing the square, we get: \begin{align*} f(x,y) &= x^2 - 6x + y^2 + 10y + 20 \\ &= x^2 - 6x + 9 + y^2 + 10y + 25 + 20 - 9 - 25 \\ &= (x - 3)^2 + (y + 5)^2 - 14 \end{align*}Notice that this function is really just a translated version of $$z = x^2 + y^2$$, so it is a paraboloid that opens up with its vertex (minimum point) at the critical point $$(3, -5)$$. We can argue that it has an absolute minimum value of $$-14$$ at the point $$(3, -5)$$, since we are adding squared terms to $$-14$$ and thus cannot get a value less than $$-14$$ for any values of $$x$$ and $$y$$, while we do obtain this minimum value of $$-14$$ at the vertex point $$(3, -5)$$.

b. Setting the partials of $$f$$ equal to $$0$$, we obtain: \begin{align*} \text{Set}\quad f_x(x,y) &= -6x -6 = 0 & \implies x &= -1 \\ \text{and}\quad f_y(x,y) &= -2y + 12 = 0 & \implies y &= 6 \end{align*}We obtain a single critical point with coordinates $$(-1, 6)$$. Next we need to determine the behavior of the function $$f$$ at this point.

To complete the square here, we first need to factor out the factors of the squared terms. Doing this and reordering the terms some gives us: \begin{align*} f(x,y) &= 12 - 3x^2 - 6x - y^2 + 12y\\ &= - 3\left(x^2 + 2x\quad\quad\right) - 1\left(y^2 - 12y \quad\quad\right) + 12 \\ &= -3\left(x^2 + 2x + 1\right) - 1\left(y^2 - 12y +36\right) + 12 +3+36\\ &= 51 - 3(x + 1)^2 - (y - 6)^2 \end{align*}Notice that this function is an elliptic paraboloid that opens down with its vertex (maximum point) at the critical point $$(-1, 6)$$. We can argue that it has an absolute maximum value of $$51$$ at the point $$(-1, 6)$$, since we are subtracting squared terms from $$51$$ and thus cannot get a value more than $$51$$ for any values of $$x$$ and $$y$$, while we do obtain this minimum value of $$51$$ at the vertex point $$(-1, 6)$$.

c. Setting the partials of $$f$$ equal to $$0$$, we obtain: \begin{align*} \text{Set}\quad f_x(x,y) &= 2x + 8 = 0 & \implies x &= -4 \\ \text{and}\quad f_y(x,y) &= -4y + 16 = 0 & \implies y &= 4 \end{align*}This gives us a critical point with coordinates $$(-4, 4)$$. To determine if $$f$$ has a local extremum or saddle point at this point, we complete the square.

Factoring out $$-2$$ from the $$y$$-squared term gives us: \begin{align*} f(x,y) &= x^2 + 8x - 2y^2 + 16y \\ &= x^2 + 8x +16 - 2\left(y^2 - 8y + 16\right) - 16 + 32 \\ &= (x + 4)^2 - 2(y - 4)^2 +16\end{align*}Since one squared term is positive and one is negative, we see that this function has the form of $$z = x^2 - y^2$$ and so it has a saddle point at its critical point. That is, $$f$$ has a saddle point at $$(-4, 4, 16)$$.

d. Setting the partials of $$f$$ equal to $$0$$, we get: \begin{align*} \text{Set}\quad f_x(x,y) &= 2x + 6y = 0 & \\ \text{and}\quad f_y(x,y) &= 6x + 2y = 0 & \implies y &= -3x \end{align*}Substituting $$-3x$$ into the first equation for $$y$$ gives us, \begin{align*}2x + 6(-3x) &= 0 \\ -16x &= 0 \\ x &= 0\end{align*}Since $$y = -3x$$, we have $$y = -3(0) = 0$$, so the critical point of $$f$$ is $$(0,0)$$. To determine the behavior of $$f$$ at this critical point, we complete the square.

\begin{align*} f(x,y) &= x^2 + 6xy + y^2 \\ &= (x^2 + 6xy + 9y^2) + y^2 - 9y^2 \\ &= (x + 3y)^2 - 8y^2 \end{align*}As this produces a difference of squares with one positive squared term and the other a negative squared term, we see that $$f$$ takes a form similar to $$z = x^2 - y^2$$ and will have a saddle point at $$(0, 0, 0)$$.

Now let's consider the quadratic approximation to a function $$z = f(x, y)$$ centered at a critical point $$(x_0, y_0)$$ of this function.

$Q(x, y) = f (x_0, y_0) + f_x(x_0, y_0) (x - x_0) + f_y(x_0, y_0) (y - y_0) + \frac{f_{xx}(x_0, y_0)}{2}(x-x_0)^2 + f_{xy}(x_0, y_0)(x-x_0)(y-y_0) + \frac{f_{yy}(x_0, y_0)}{2}(y-y_0)^2$

But, since the point $$(x_0, y_0)$$, in this case, is a critical point of $$f$$, we know that $$f_x(x_0, y_0) = 0$$ and $$f_y(x_0, y_0) = 0$$.

This allows us to simplify $$Q(x, y)$$ to just:

$Q(x, y) = f (x_0, y_0) + \frac{f_{xx}(x_0, y_0)}{2}(x-x_0)^2 + f_{xy}(x_0, y_0)(x-x_0)(y-y_0) + \frac{f_{yy}(x_0, y_0)}{2}(y-y_0)^2$

Now we need to complete the square on this quadratic polynomial in two variables to learn how we can classify the behavior of this function at this critical point. Remember that the original function will share the same behavior (max, min, saddle point) as this 2nd-degree Taylor polynomial at this critical point.

To make this process easier, let's make some substitutions. Let's choose to let $$u = x - x_0$$ and $$v = y - y_0$$,

and let \begin{align*} a &= \frac{f_{xx}(x_0, y_0)}{2}, \\ b &= f_{xy}(x_0, y_0), \\ c &= \frac{f_{yy}(x_0, y_0)}{2} \,\text{and} \\ d &= f (x_0, y_0) \end{align*}

Then we need to complete the square on the polynomial: $Q(x,y) = au^2 +buv + cv^2 + d$

Completing the square:

First we factor out the coefficient of $$u^2$$: $= a\left[ u^2 + \frac{b}{a}uv + \frac{c}{a}v^2\right] + d$

Next, we complete the square using the first two terms: $= a\left[ \left(u^2 + \frac{b}{a}uv + \left(\frac{b}{2a}v\right)^2\right) + \frac{c}{a}v^2 - \left(\frac{b}{2a}v\right)^2 \right] + d$

Rewriting the perfect square trinomial as the square of a binomial and combining the $$v^2$$ terms yields:

\begin{align*} &= a\left[ \left(u+ \frac{b}{2a}v\right)^2 + \left(\frac{c}{a} - \frac{b^2}{4a^2}\right)v^2 \right] + d \\ &= a\left[ \left(u+ \frac{b}{2a}v\right)^2 + \left(\frac{4ac}{4a^2} - \frac{b^2}{4a^2}\right)v^2 \right] + d \\ &= a\left[ \left(u+ \frac{b}{2a}v\right)^2 + \left(\frac{4ac-b^2}{4a^2}\right)v^2 \right] + d \end{align*}

Note that the shape of this function's graph depends on the sign of the coefficient of $$v^2$$. And the sign of this coefficient is determined only by its numerator, as the denominator is always positive (being a perfect square). This expression, $$4ac-b^2$$, is called the discriminant, as it helps us discriminate (tell the difference between) which behavior the function has at this critical point.

If $$D = 4ac-b^2\gt 0$$, then the two squared terms inside the brackets are both positive, and

• if $$a = \frac{f_{xx}(x_0, y_0)}{2} \gt 0$$, the function $$f$$ opens upwards with a local minimum at the critical point $$(x_0, y_0)$$. Note it would be similar to the form, $$z = x^2 + y^2$$.
• if $$a = \frac{f_{xx}(x_0, y_0)}{2} \lt 0$$, the function $$f$$ opens downwards with a local maximum at the critical point $$(x_0, y_0)$$. Note it would be similar to the form, $$z = -\left(x^2 + y^2\right)$$.

If $$D = 4ac-b^2 \lt 0$$, then either

• the two squared terms inside the brackets have opposite signs (meaning $$f$$ is concave up along a line parallel to the $$x$$-axis and concave down along a line parallel to the $$y$$-axis, or vice-versa) or
• the $$b^2$$ term, representing the square of the mixed partial $$f_{xy}(x_0, y_0)$$, is larger than the positive product of the two 2nd-partials $$f_{xx}(x_0, y_0)$$ and $$f_{yy}(x_0, y_0)$$. This means that even if the surface is concave up in both $$x$$- and $$y$$-directions, or concave down in both $$x$$- and $$y$$-directions, a large mixed partial can offset these and cause the surface to have a saddle point at the point $$(x_0, y_0)$$.

In either case, the quadratic polynomial will be in the form of $$z = x^2 - y^2$$ or $$z = y^2 - x^2$$ (i.e., it will be the difference of two squared terms), so we get a saddle point at the critical point $$(x_0, y_0)$$.

But if $$D = 4ac-b^2 = 0$$, the quadratic polynomial reduces to $$Q(x,y) = a\left(u+ \frac{b}{2a}v\right)^2 + d$$, whose graph is a parabolic cylinder, so the behavior of the function is not clear at the critical point $$(x_0, y_0)$$.

Now remembering the values of the constants $$a$$, $$b$$, and $$c$$ from above, we see that: \begin{align*} D(x_0, y_0) &= 4\frac{f_{xx}(x_0, y_0)}{2}\frac{f_{yy}(x_0, y_0)}{2} - \big(f_{xy}(x_0, y_0)\big)^2 \\ &= f_{xx}(x_0, y_0)f_{yy}(x_0, y_0) - \big(f_{xy}(x_0, y_0)\big)^2 \end{align*}

This formula is called the Second Partials Test, and it can be used to classify the behavior of any function at its critical points, as long as its second partials exist there and as long as the value of this discriminate is not zero.