# Section 1.4 Question 1

## What is a linear function of several independent variables?

In section 1.1, we introduced a linear function of one variable, y = mx + b. There was nothing special about the names of the variables, x and y, or the names of the constants, m and b. Another possible form for a linear function of x and y is y = a0 + ax. In this format, a0 is the vertical intercept and a1 is the slope.

When several independent variables are introduced, it is prudent to use names for the variables that make sense. If one variable is named x, we can extend this to n variables using subscripts. Subscripts are numbers that appear to the right of the variable and slightly lowered. The subscript is a part of the variable’s name and is useful to show generically that there are many variables. For instance, if we wanted to define a function with three independent variables that describe the quantities of three different products, we might use Q1, Q2, and Q3.

In general, let x1, x2, … , xn be the names of n independent variables.

A linear function of n independent variables x1x2, … , xn is any equation that can be written in the form

z = a0 + a1 x1 + a2 x2 + ··· + an xn

In this form, we say that z is a linear function of x1x2, … , xn. The letters a0, a1, … , an are real numbers corresponding to constants.

Function notation applies to functions of several independent variables as well as functions of one independent variable. Recall that a linear function of one variable x named f  would be written as f (x) = a0 + ax. The independent variable for the function is placed in parentheses after the name to distinguish the variables from the constants. For a linear function of n independent variables, the n independent variables are placed in the parentheses after the name to give

f (x1x2, … , xn) = a0 + a1 x1 + a2 x2 + ··· + an xn

### Example 1      Find Function Values

If f (x1x2x3) = 10 – 2x1 + x2+ 3x3, find the value of (6, -1, 2).

Solution  Substitute x1 = 6, x2 = -1, and x3 = 2 into the function to yield

f (6, -1, 2) = 10 – 2(6) + (-1)+ 3(2) = 3