Multivariable-Calculus · Unit 9 · Video 1 · Interactive Practice

Classifying Critical Points: Completing the Square on ax² + bxy + cy²

IKey Formulas

Formula	Name	What it says
$w = a\left(x + \frac{b}{2a}y\right)^2 + \left(c - \frac{b^2}{4a}\right)y^2$	Completing the square ( $a \neq 0$ )	The $bxy$ term is absorbed into one square
$w = \frac{1}{4a}\left[4a^2\left(x + \frac{b}{2a}y\right)^2 + (4ac - b^2)\,y^2\right]$	Two-squares form	The sign of $4ac - b^2$ decides sum or difference of squares
$w_{xx} = 2a,\quad w_{xy} = b,\quad w_{yy} = 2c$	Second derivatives of $w$	The coefficients are second derivatives at the origin

Key Insight: At the critical point $(0, 0)$ of $w = ax^2 + bxy + cy^2$ : $4ac - b^2 < 0$ gives a saddle; $4ac - b^2 = 0$ is degenerate; $4ac - b^2 > 0$ gives a minimum if $a > 0$ and a maximum if $a < 0$ .

IIVisualization 1 — Three Coefficients, One Verdict

The sign of $4ac - b^2$ decides whether the level curves are ellipses, hyperbolas or parallel lines.

a 1 b −2 c 3

IIIVisualization 2 — Which Square Wins?

With $u = x + \frac{b}{2a}y$ and $v = y$ , the sign of $w$ at a point depends on which square term is larger.

Function

IVVisualization 3 — Every Line Through the Origin

At a minimum every line through the origin cuts out an upward parabola; at a saddle some open downward.

Function Direction θ 180°

💡 A degenerate saddle such as $x^3 + y^3$ needs higher-order terms: with quadratic terms alone, a degenerate point is a degenerate minimum or maximum, depending on the sign of $a$ .

VQuiz Questions

Problem 1 · Classify the Origin

Given: $w = 2x^2 + 3xy + 2y^2$ — what type of critical point is the origin?

✅ Correct!

4ac - b^2 = 16 - 9 = 7 > 0

and

a = 2 > 0

, so

w

is a positive multiple of a sum of two squares.

❌ Not quite. When

4ac - b^2 > 0

the bracket is never negative, and the factor

\frac{1}{4a}

decides the sign: here

a = 2 > 0

❌ Not quite. Degenerate means

4ac - b^2 = 0

exactly; here

4ac - b^2 = 4 \cdot 2 \cdot 2 - 3^2 = 7

❌ Not quite. Compute

4ac - b^2

with

a = 2

b = 3

c = 2

and check its sign.

Show solution

Here $a = 2$ , $b = 3$ , $c = 2$ , so

4ac - b^2 = 4(2)(2) - 3^2 = 16 - 9 = 7 > 0.

Completing the square:

w = 2\left(x + \tfrac{3}{4}y\right)^2 + \left(2 - \tfrac{9}{8}\right)y^2 = 2\left(x + \tfrac{3}{4}y\right)^2 + \tfrac{7}{8}y^2.

Both terms are $\ge 0$ and $w(0, 0) = 0$ , so the origin is a local minimum (in fact the global minimum).

Problem 2 · Positive Coefficients Are Not Enough

Given: $w = x^2 + 6xy + 4y^2$ — what type of critical point is the origin?

✅ Correct!

4ac - b^2 = 16 - 36 = -20 < 0

: the large cross term makes

w

a difference of two squares.

❌ Close, but… the cross term matters. Along the line

x = -3y

you get

w = 9y^2 - 18y^2 + 4y^2 = -5y^2 < 0

, while along the

x

-axis

w = x^2 > 0

❌ Not quite. Compute

4ac - b^2

with

a = 1

b = 6

c = 4

Show solution

With $a = 1$ , $b = 6$ , $c = 4$ :

4ac - b^2 = 16 - 36 = -20 < 0.

Completing the square:

w = (x + 3y)^2 + (4 - 9)y^2 = (x + 3y)^2 - 5y^2,

a difference of two squares. On the $x$ -axis ( $y = 0$ ) $w = x^2 > 0$ ; on the line $x + 3y = 0$ $w = -5y^2 < 0$ . The origin is a saddle point, even though $a > 0$ and $c > 0$ .

Problem 3 · Complete the Square, Then Decide

Given: $w = 2x^2 - 8xy + 5y^2$ — write $w$ in completed-square form and classify the origin.

Completed-square form

Type of the origin

✅ Correct!

w = 2(x - 2y)^2 - 3y^2

is a difference of two squares, matching

4ac - b^2 = 40 - 64 = -24 < 0

: a saddle.

❌ Check the shift. The square is

\left(x + \frac{b}{2a}y\right)^2

with

\frac{b}{2a} = \frac{-8}{4} = -2

, not

\frac{b}{a} = -4

2(x - 4y)^2

has cross term

-16xy

❌ Check the $y^2$ term.

2(x - 2y)^2 = 2x^2 - 8xy + 8y^2

already contains

8y^2

, so only

5 - 8 = -3

y^2

is left over.

❌ Check the $y^2$ term. The correction is

\frac{b^2}{4a} = \frac{64}{8} = 8

, not

\frac{b^2}{4} = 16

❌ Check your algebra. Factor

2

out of

2x^2 - 8xy

, then complete the square on

x^2 - 4xy

❌ Check the type. One square has a positive coefficient and the other a negative one.

Show solution

Here $a = 2$ , $b = -8$ , $c = 5$ . Factor $a$ out of the $x$ -terms:

w = 2\left(x^2 - 4xy\right) + 5y^2.

Since $(x - 2y)^2 = x^2 - 4xy + 4y^2$ , we have $2(x - 2y)^2 = 2x^2 - 8xy + 8y^2$ , so

w = 2(x - 2y)^2 + (5 - 8)y^2 = 2(x - 2y)^2 - 3y^2.

The same numbers from the formula: $\frac{b}{2a} = -2$ and $c - \frac{b^2}{4a} = 5 - \frac{64}{8} = -3$ .

One square with coefficient $+2$ , one with $-3$ : a difference of squares, so the origin is a saddle point. Check: $4ac - b^2 = 40 - 64 = -24 < 0$ .

Problem 4 · A Coefficient to Tune

Given: $w = kx^2 + 6xy + 3y^2$ — for which values of $k$ is the origin a local minimum?

✅ Correct! A minimum needs

4ac - b^2 = 12k - 36 > 0

and

a = k > 0

; both hold exactly when

k > 3

❌ Not quite.

a > 0

alone is not enough: for

k = 1

4ac - b^2 = 12 - 36 < 0

and the origin is a saddle.

❌ Close, but… at

k = 3

4ac - b^2 = 0

w = 3(x + y)^2

is zero along the whole line

x = -y

, a degenerate critical point.

❌ Not quite. The test quantity is

4ac - b^2 = 4 \cdot k \cdot 3 - 36

, not

ac - b^2

❌ Not quite. Write

4ac - b^2

as a function of

k

and require it to be positive with

a > 0

Show solution

Here $a = k$ , $b = 6$ , $c = 3$ . The origin is a minimum when $4ac - b^2 > 0$ and $a > 0$ :

4ac - b^2 = 4(k)(3) - 6^2 = 12k - 36 > 0 \iff k > 3.

Every $k > 3$ is also positive, so the condition is $k > 3$ .

At $k = 3$ : $w = 3x^2 + 6xy + 3y^2 = 3(x + y)^2$ , degenerate. For $k < 3$ : $4ac - b^2 < 0$ , a saddle.

In terms of second derivatives, $w_{xx} = 2k$ , $w_{xy} = 6$ , $w_{yy} = 6$ , and $4ac - b^2 = w_{xx}w_{yy} - w_{xy}^2 = 12k - 36$ .

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