To find the [tex]$x$[/tex]-intercepts and the [tex]$y$[/tex]-intercept of the function [tex]\( f(x) = (x-3)(x-1)(x+3) \)[/tex], we'll proceed step-by-step.
### Step 1: Finding the [tex]$x$[/tex]-Intercepts
The [tex]$x$[/tex]-intercepts occur where the function [tex]\( f(x) \)[/tex] equals zero. That is, we need to solve the equation:
[tex]\[
(x-3)(x-1)(x+3) = 0
\][/tex]
For the product of these factors to equal zero, at least one of the factors must be zero. Therefore, we solve each factor individually:
[tex]\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\][/tex]
[tex]\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\][/tex]
[tex]\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\][/tex]
Thus, the [tex]$x$[/tex]-intercepts are at the points [tex]\((3, 0)\)[/tex], [tex]\((1, 0)\)[/tex], and [tex]\((-3, 0)\)[/tex].
### Step 2: Finding the [tex]$y$[/tex]-Intercept
The [tex]$y$[/tex]-intercept occurs where [tex]\( x = 0 \)[/tex]. We need to evaluate the function [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = (0-3)(0-1)(0+3)
\][/tex]
Calculating inside the parentheses:
[tex]\[
f(0) = (-3)(-1)(3)
\][/tex]
Multiplying these values together:
[tex]\[
f(0) = (-3) \times (-1) \times 3 = 9
\][/tex]
Therefore, the [tex]$y$[/tex]-intercept is at the point [tex]\((0, 9)\)[/tex].
### Summary
The [tex]$x$[/tex]-intercepts and [tex]$y$[/tex]-intercept of the function [tex]\( f(x) = (x-3)(x-1)(x+3) \)[/tex] are:
- [tex]$x$[/tex]-intercepts: [tex]\((-3, 0)\)[/tex], [tex]\((1, 0)\)[/tex], [tex]\((3, 0)\)[/tex]
- [tex]$y$[/tex]-intercept: [tex]\((0, 9)\)[/tex]
These are the coordinate pairs where the function intersects the respective axes.
