Linear Functions
In single-variable calculus, we studied linear equations because they allowed us to
approximate complicated curves with a fairly simple linear function. Our goal will be to
use linear functions in two dimensions to approximate complicated surfaces.
A linear function in one variable could be written in point-slope form. That is, we said
that y = y0 + m(x – x0). Once we knew a point on the linear and the slope, we had all the
information necessary to specify the equation.
When dealing with two variables, we have a similar point-slope form. This time,
however, we have two slopes: one in the x-direction and another in the y-direction.
Equation of a Linear Function
If a linear function (a plane) passes through the point (x0, y0, z0) and has slope
m in the x-direction and slope n in the y-direction, the equation of the linear
function (plane) is given by:
z = f(x, y) = z0 + m(x – x0) + n(y – y0)
Letting c = z0 – mx0 – ny0, we can also write f(x, y) in the form
z = f(x, y) = c + mx + ny
Just as two points uniquely determine the equation of a line in 2-space, three points
uniquely determine the equation of a plane in 3-space.
Example 1:
Find the equation of the plane passing through the points (1, 2, 3), (1, 4, 1), (2, 2, 5).
Solution:
Notice that the first two points have the same x-value. So, we can use them to find the
slope of the plane in the y-direction. As y changes from 2 to 4, the z-value changes from 3
to 1. Thus, the slope in the y-direction is given by n = Dz/Dy = (1 – 3)/(4 – 2) = -1.
The first and third points share the same y-value. Thus, we can use them to find the
slope of the plane in the x-direction. As x changes from 1 to 2, the z-value changes from 3
to 5. Thus, the slope in the x-direction is given by m = Dz/Dx = (5 – 3)/(2 – 1) = 2.
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Because the plane passes through the point (1, 2, 3), we have that its equation is given
by z = 3 + 2(x – 1) – (y – 2) = 3 + 2x – y. A graph of the plane appears below.
Figure 1: Graph of z = f(x, y) = 3 + 2x – y
We notice that for a fixed value of y, the graph slopes upwards (in accordance with the
fact that it has a slope of +2 in the x-direction) and for a fixed value of x, the graph slopes
downward (in accordance with the fact that it has a slope of -1 in the y-direction).
Recall when we first introduced functions of two variables, we presented them in
terms of a table. There are some key characteristics that we can look for in a table of
values that allow us to determine if we are dealing with a linear function.
Tables of Values of Linear Functions
A table of values of a linear function satisfies the following two properties:
1. All the columns have the same slope
2. All the rows have the same slope (usually diff. from the slope of the cols)
Example 2:
Show that the following table corresponds to a linear function. Find its equation.
x
2
0
2
y
1
3
1
1
2
0
6
4
2
1
9
7
5
Solution:
Notice that for a fixed x-value, the function increases by 3 for every one unit increase
in y. That is, we have that n = Dz/Dy = 3/1 = 3. Also, for a fixed y-value, the function
decreases by 2 for every two unit increase in x. That is, m = Dz/Dx = -2/2 = -1. Also, we
observe that the function passes through the point (0, 0, 4).
Using the second form of the equation of a linear function, we have that the equation
of the linear function is given by z = 4 – x + 3y.
Example 3:
The following table contains values from a linear function. Fill in the blank and
determine the equation of the linear function.
y
x
1
2
14
3
_
4
24
2
_
_
26
Solution:
Notice that for a fixed y-value, as x increases from 1 to 2 (Dx = 1), the z-value
increases by 2. This tells us that m = Dz/Dx = 2/1 = 2. From this, we can deduce that the
lower-left entry of the table is 16.
Notice, also, that for a fixed x-value, as y increases from 2 to 4 (Dy = 2), the z-value
increases by 10. This tells us that the n = Dz/Dy = 10/2 = 5. From this, we see that the
middle entry of the top row will be 14 + 5 = 19 and the middle entry of the bottom row
will be 16 + 5= 21. Thus, the completed table is as follows:
y
x
1
2
2
14
16
3
19
21
4
24
26
We have found that m = 2 and n = 5. Using the fact that the function passes through
the point (1, 2, 14), we can use the first form of the equation of a linear function to see
that the function is given by z = f(x, y) = 14 + 2(x – 1) + 5(y – 2) = 2 + 2x + 5y.
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In the previous section, we gave an example where the contour diagram of a linear
function was specified and the reader was asked to graph the equation.
All linear functions have contour diagrams that share some key properties. We list
them below.
Contour Diagrams of Linear Functions
Contour diagrams of linear functions satisfy the following two properties:
1. The level curves are parallel straight lines
2. Each of the level curves are evenly spaced
Example 4:
Below are three contour diagrams. Which one(s) correspond to a linear function? For
those that do not, what property is violated?
y
y
x
I
y
x
II
x
III
Solution:
Graph I cannot correspond to a linear function since the level curves are not parallel
straight lines. The contours, however, are evenly spaced. This actually corresponds to the
graph of the function f ( x, y )  x 2  y 2 .
Graph II does correspond to a linear function. The level curves are parallel straight
lines and the contours are evenly spaced.
Graph III cannot correspond to a linear function since the level curves are not evenly
spaced. The contours are parallel straight lines, though.
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It is also possible to determine the equation of the linear function by examining the
contour diagram.
Example 5:
Find the equation of the linear function whose contour diagram is below.
Solution:
The equation of the linear function is given by z = z0 + m(x – x0) + n(y – y0). Notice
that when x = 0 and y = 0, we have that z = 0. So, we have the point (x0, y0, z0) = (0, 0, 0)
on the equation.
As we move out one unit in the positive y-direction, the function value decreases by 2.
This implies that n = -2. As we move out two units in the x-direction, the function value
increases by 2. This implies that m = 2/2 = 1.
So, the equation of our linear function is given by z = 0 + 1(x – 0) – 2(y – 0) = x – 2y.
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