8.3 Quadratic Functions
In this section we shall look at the connection between quadratic functions and symmetric
matrices.
Definition 1. A quadratic function z = f(x, y) of two variables x and y is one of the form
z = f(x, y) = ax2 + 2bxy + cy2 + dx + ey + f
(1)
where a, b, c, d, e and f are fixed numbers. A quadratic form (or homogeneous quadratic
function) is a quadratic function where only the x2, xy and y2 terms are present, i.e. one of
the form
z = f(x, y) = ax2 + 2bxy + cy2
(2)
Example 1. The following are quadratic functions.
z = x2 + 4y2
z = x2 - 4y2
z = 5x2 + 20y2
z = 17x2 - 12xy + 8y2
z = 17x2 - 12xy + 8y2 + 2x - 6y + 3
The first four are quadratic forms.
A quadratic form can be written in vector form using a symmetric matrix.
Proposition 1.
(3)
a b
x
ax2 + 2bxy + cy2 = (x, y)  b c   y  = uTQu
(4)
a b
x
x
ax2 + 2bxy + cy2 + dx + ey + f = (x, y)  b c   y  + (d, e)  y  + f = uTQu + pu + f
x
a b
where u =  y , Q =  b c  and p = (d, e).
Proof. The first part follows from the fact that
z = ax2 + 2bxy + cy2 = (ax2 + bxy) + (bxy + cy2) = x(ax + by) + y(bx + cy)
ax + by
a b
x
= (x, y)  bx + cy  = (x, y)  b c   y  = uTQu
8.3 - 1
The second part follows easily from the first. //
Example 2.
17x2 – 12xy + 8y2 = (17x2 - 6xy) + (- 6xy + 8y2) = x(17x - 6by) + y(- 6x + 8y)
17x - 6y
17 - 6
x
= (x, y)  - 6x + 8y  = (x, y)  - 6 8   y  = uTQu
Problem 1. Write 6x2 – 4xy + 3y2 in the form (3). Answer: Q =  - 2
6 -2
3 .
Definition 2. The graph of a function z = f(x, y) is the set of all points
(x, y, z) = (x, y, f(x, y)) in space where the z coordinate is obtained by applying the
function to the x and y coordinates. A level curve of z = f(x, y) is the graph of the
equation f(x, y) = k for some fixed number k. The contour map of z = f(x, y) is the set of
all level curves.
The graph will be a surface in three dimensions while each level curve is a curve in the
xy-plane. A graph has an infinite number of non-intersecting level curves, one for each
number k. The contour map consists of all these level curves taken together and it
divides up the plane into these non-intersecting curves. Often one can get a good picture
of the graph from the contour map. Many questions involving the function can be
visualized by means of the graph and/or contour map. In this section we see how to make
the contour map and graph of quadratic functions.
We begin with quadratic functions of the form
(5)
x2
y2
z = p 2 + q2
x2
y2
where p and q are fixed numbers. In this case the level curves have the form p2 + q2 = k.
If k < 0 there are no x and y satisfying the equation since the left side is positive and the
right side is negative, so there is no level curve for negative values of k. If k = 0 the only
pair of values of x and y satisfying this equation is (x, y) = (0, 0), so the level curve is just
a single point, i.e. the origin. If k > 0 then we divide by k and rewrite the level curves as
(
x2
k p)2
+
(
y2
k q)2
= 1
or
x2
a2
y2
+ b2 = 1
with a = k p and b = k q. So the level curve is an ellipse extending from – k p to k p
on the x axis and – k q to k q on the y axis. So the level curves are concentric ellipses
x2
y2
which get larger as k increases. The resulting surface z = p2 + q2 is bowl shaped
opening upward with its low point at the origin. It is called an elliptic paraboloid with its
8.3 - 2
vertex at the origin. The curves obtained by fixing y and letting x vary are parabolas as
are the curves obtained by fixing x and letting y vary.
Example 3. Consider the function z = x2 + 4y2. Draw the level curves for z = 4 and z = 8
and the graph of the function.
x2
y2
The level curve x2 + 4y2 = 4 for z = 4 is the ellipse 22 + 12 = 1 extending from – 2 to 2 on
the x axis and - 1 to 1 on the y axis. The level curve
x2 + 4y2 = 8 for z = 8 is the ellipse
x2
(2 2)2
+
(
y2
2 )2
2
1
= 1 extending from – 2 2 to 2 2 on
4
2
2
the x axis and - 2 to 2 on the y axis. These are
shown at the right along with the graph of
z = x2 + 4y2.
4
1
2
The graph of a quadratic functions of the form
x
y
z = - p2 + q2
2
(6)
2
is obtained by reflecting the graph of (5) across
the z axis. It is an elliptic paraboloid openting
down with its vertex at the origin.
Now consider quadratic functions of the form
x2
(7)
y2
z = p 2 - q2
x2
y2
where p and q are fixed numbers. In this case the level curves have the form p2 + q2 = k.
If k > 0 then when we divide by k the level curves have the form
(
x2
k p)2
-
(
y2
k q)2
= 1
x2
a2
or
y2
- b2 = 1
with a = k p and b = k q. So the level curve is a hyperbola with vertices at – k p and
kp
b
b
on the x axis and asymptotes y = a x and y = - a x. So the level curves are hyperbolas
b
which get larger as k increases. If k = 0 then the level curves are the straight lines y = a x
b
and y = - a x. If k < 0 then when we divide by k the level curves have the form
(
y2
- k q)2
+
(
x2
- k p)2
= 1
or
8.3 - 3
y2
b2
x2
- a2 = 1
with a = - k p and b = - k q. So the level curve is a hyperbola with vertices at – k q and
kq
b
b
on the y axis and asymptotes y = a x and y = - a x. Again these level curves are
hyperbolas which get larger as k decreases.
x2
y2
The resulting surface z = p2 - q2 is saddle shaped surface called an hyperbolic
paraboloid. It is said to open up along the x axis and open down along the y axis. The
curves obtained by fixing y and letting x vary are parabolas as are the curves obtained by
fixing x and letting y vary.
Example 4. Consider the function z = x2 - 4y2. Draw the level curves for z = 4, z = 8 and
z = - 4 and the graph of the function.
2
The level curve x2 - 4y2 = 4 for z = 4 is the hyperbola
x2 y2
22 12
asymptotes y = x/2 and y = - x/2. The level curve
x2 - 4y2 = 8 for z = 8 is the hyperbola
x
(2
1
= 1 with vertices at – 2 and 2 on the x axis and
2
2)2
-
y
2
( 2)2
4
= 1 with vertices – 2 2 to 2 2 on the
2
2
4
1
2
x axis and same asymptotes. The level curve
y2
x2
x2 - 4y2 = - 4 for z = - 4 is the hyperbola 12 - 22 = 1
with vertices at – 2 and 2 on the y axis and same
asymptotes. These are shown at the right along with
the graph of z = x2 - 4y2.
The graph of quadratic functions of the form
(8)
y2
x2
z = q 2 - p2
is similar to the graph of (7) only it opens up
along the y axis and opens down along the x axis.
We can summarize the above discussion by
means of the following proposition.
Proposition 2. Let z = ax2 + cy2. If a and c are both positive the graph is an elliptic
parabolid opening up with its vertex at the origin. If a and c are both negative the graph
is an elliptic parabolid opening down with its vertex at the origin. If a is positive and c is
negative the graph is a hyperbolic parabolid that opens up along the x axis and opens
down along the y axis. If c is positive and a is negative the graph is a hyperbolic
parabolid that opens up along the y axis and opens down along the x axis.
For general quadratic forms we have the following proposition.
8.3 - 4
Proposition 3. Let
(9)
a b
x
z = ax2 + 2bxy + cy2 = (x, y)  b c   y  = uTQu
x
a b
where u =  y , Q =  b c . Let 1 and 2 be the eigenvalues and v1 and v2 be the
eigenvectors of Q. If we choose a new coordinate system for the xy plane with r axis
through v1 and s axis through v2 then in this new coordinate system the equation (9)
becomes z = 1r2 + 2s2. In particular,
(1) The level curve ax2 + 2bxy + cy2 = k becomes the level curve 1r2 + 2s2 = k in
the rs coordinate system.
(2) If 1 and 2 are both positive the graph of (9) is an elliptic parabolid opening up
with its vertex at the origin. If 1 and 2 are both negative the graph is an
elliptic parabolid opening down with its vertex at the origin. If 1 is positive
and 2 is negative the graph is a hyperbolic parabolid that opens up along the r
axis and opens down along the s axis. If 2 is positive and 1 is negative the
graph is a hyperbolic parabolid that opens up along the s axis and opens down
along the r axis.
Proof. Suppose we have normalized the eigenvectors as discussed in the previous
section. Then Q = TDT-1 where T is the matrix whose columns are v1 and v2 and
 0
D =  01  . We substitute this into z = uTQu to get z = uTTDT-1u = (TTu)TDT-1u. As
2
discussed in the previous section T is orthogonal so T-1 = TT. So z = (T-1u)TDT-1u.
We can express a vector u in the plane as a linear combination of the eigenvectors v1 and
v2, i.e. u = rv1 + sv2 where r and s are the coordinates of u with respect to this new
r
r
coordinate system. Algebraically we have u = T  s . So  s  = T-1u. So
r T
r
z =  s  D  s  = 1r2 + 2s2. //
17 - 6
x
Example 5. Consider the quadratic form z = 17x2 - 12xy + 8y2 = (x, y)  - 6 8   y  =
uTQu. (a) Sketch the level curve 17x2 - 12xy + 8y2 = 20. (b) What kind of surface is the
graph of z = 17x2 - 12xy + 8y2. (c) Sketch the graph of z = 17x2 - 12xy + 8y2.
By example 3 in section 6.1 the eigenvalues of Q are 1 = 5 and 2 = 20 and the
1
-2
eigenvectors are v1 =  2  and v2 =  1 . Let the r and s axes be the lines through v1 and
v2. The level curve 17x2 - 12xy + 8y2 = 20 becomes 5r2 + 20s2 = 20 in the rs coordinate
8.3 - 5
system. This is r2 + 4s2 = 4. It is an ellipse and its is below. The graph of z = 17x2 - 12xy
+ 8y2 is an elliptic paraboloic opening up and its graph is also below.
2
1
2
1
1
2
1
2
6 -2
x
Problem 2. Consider the quadratic form z = 6x2 – 4xy + 3y2 = (x, y)  - 2 3   y  = uTQu.
(a)
Find a new coordinate system for the xy plane such that in this new coordinate system
r
the graph of the function z = 6x2 – 4xy + 3y2 has the form z = Ar2 + Bs2 where  s  are
x
the coordinates of  y  relative to this new coordinate system. Specify this new
coordinate system by giving vectors along the new coordinate axes. Also, give A and
B and the angle which the r axis makes with respect to the x axis.
(b)
Sketch the level curve 6x2 – 4xy + 3y2 = 98 showing relevant lengths. What kind of
curve is it.
(c)
Sketch the graph of z = 6x2 – 4xy + 3y2. What kind of surface is it.
1
-2
Answers: The r axis is through v1 =  2  and the s axis is through v2 =  1 . In this new
coordinate system the function z = 6x2 – 4xy + 3y2 becomes z = 2r2 + 7s2. The angle between
the r axis and x axis is tan-1(2)  1.11 radians  63.4. The level curve 6x2 – 4xy + 3y2 = 98 is
2r2 + 7s2 = 98 in the rs coordinate system. It is an ellipse whose axis goes from – 7 to 7 on
the r axis and - 14 to 14 on the s axis. The graph of z = 6x2 – 4xy + 3y2 is a elliptic
paraboloid.
5 3
x
Example 6. Consider the quadratic form z = 5x2 + 6xy - 3y2 = (x, y)  3 - 3   y  = uTQu.
(a) Sketch the level curve 5x2 + 6xy - 3y2 = 24. (b) What kind of surface is the graph of
z = 5x2 + 6xy - 3y2. (c) Sketch the graph of z = 5x2 + 6xy - 3y2.
First, find the eigenvalues and eigenvectors of Q.
8.3 - 6
5-
3
0 = det( Q - I ) =  3 - 3 -   = (5 - )(- 3 - ) – 9
= 2 - 2 - 24 = ( - 6)( + 4)
x
So the eigenvalues are 1 = 6 and 2 = - 4. An eigenvector v =  y  for 1 = 6 satisfies
 0  = (A - 6I)v =  - 1 3   x 
0
 3 -9 y
So
- x + 3y = 0
3x - 9y = 0
3
The solution to both equations is x = 3y, so v1 =  1  is an eigenvector for 1 = 6. Since
-3
the eigenvectors are orthogonal an eigenvector for 2 = -4 is v2 =  1 . Let the r and s
axes be the lines through v1 and v2. The level curve 5x2 + 6xy - 3y2 = 24 becomes
r2
6r2 - 4s2 = 24 in the rs coordinate system. This is 22 -
(
s2
6)2
= 1. It is a hyperbola opening
up the r axis with its vertices at – 2 and 2 on the r axis and asymptotes s = 6r/2 and
s = - 6r/2. Its graph is below. The graph of z = 5x2 + 6xy - 3y2 is an hyperbolic
paraboloic opening up along the r axis and opening down along the s axis and its graph is
also below.
6
4
2
6
4
2
2
4
6
2
4
Finally, we consider general quadratic
functions of the form (1). They can be
reduced to quadratic forms of the form (2) by creating a new coordinate system parallel
x
to the orignal but with the origin shifted to a point u0 =  y00 .
6
8.3 - 7
Proposition 4. Let
(10)
a b
x
x
z = ax2 + 2bxy + cy2 + dx + ey + f = (x, y)  b c   y  + (d, e)  y  + f = uTQu + pu + f
x
a b
x
1
where u =  y , Q =  b c . Assume Q is invertible and  y00  = - 2 Q-1pT. If we choose a
x
new coordinate system for the xy plane with origin  y00  and r and s axes parallel to the x
x
and y axes then the equation (9) becomes z = vTQv + (f + pq + qTQq) where q =  y00  and
v = u – q.
Proof. One has u = v + q. Substitute this into z = uTQu + pu + f. We get
z = (v + q)TQ(v + q) + p(v + q) + f
= vTQv + vTQq + qTQv + qTQq + pv + pq + f
vTQq is a number so vTQq = (vTQq)T = qTQTv = qTQv since Q is symmetric. So
z = vTQv + (2qTQ + p)v + (f + pq + qTQq)
1
Since q = - 2 Q-1pT it follows that 2Qq = - pT or - p = 2qTQT = 2qTQ since Q is symmetric.
So z = vTQv + (f + pq + qTQq). //
Example 7. Consider the quadratic form
(11)
z = 17x2 - 12xy + 8y2 - 10x - 20y + 15
17 - 6
x
x
= (x, y)  - 6 8   y  + (- 10, - 20)  y  + 15
= uTQu + pu + f
(a) Sketch the level curve 17x2 - 12xy + 8y2 - 10x - 20y + 15 = 10. (b) What kind of
surface is the graph of z = 17x2 - 12xy + 8y2 - 10x - 20y + 15. (c) Sketch the graph of
z = 17x2 - 12xy + 8y2 - 10x - 20y + 15.
-1
1
1 17 - 6
- 10
1 8 6
1
1 20
1
One has q = - 2 Q-1pT = - 2  - 6 8   - 20  = 20  6 17  2  = 20  42  =  2  and
1
17 - 6
1
5
f + pq + qTQq = 15 + (- 10, - 20)  2  + (1, 2)  - 6 8   2  = 15 – 50 + (1, 2)  10  =
x
1
15 – 50 + 25 = - 10. The rs coordinate system the has origin at  y  =  2  and axes
parallel to the x and y axes. In this coordinate system z = 17r2 – 12rs + 8s2 - 10 and the
level curve z = 10 becomes 17r2 – 12rs + 8s2 = 20. So we take the level curve
8.3 - 8
17x2 - 12xy + 8y2 = 20 that we made in Example 5 and move it to the right one unit and
1
up two units. So the level curve is an ellipse with center at  2  and it is shown below.
The graph z = 17x2 - 12xy + 8y2 - 10x - 20y + 15 is an elliptic paraboloid. To make it we
take the graph of z = that we made in Example 5 and move it one unit in the x direction,
2 units in the y direction and – 10 units in the z direction. It is also shown below.
4
3
2
1
1
1
2
3
1
8.3 - 9