To understand the relationship between the real zero(s) and [tex]\( x \)[/tex]-intercept(s) of the function [tex]\( f(x) = \frac{3x(x-1)}{x^2(x+3)(x+1)} \)[/tex], we need to analyze the numerator and denominator separately.
1. Finding the Real Zeros:
To find the real zeros (or roots) of the function, we need to set the numerator equal to zero:
[tex]\[
3x(x - 1) = 0
\][/tex]
Solving this equation, we get:
[tex]\[
3x = 0 \quad \text{or} \quad x - 1 = 0
\][/tex]
Which simplifies to:
[tex]\[
x = 0 \quad \text{or} \quad x = 1
\][/tex]
Therefore, the potential real zeros of the function are [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex].
2. Identifying the Domain Restrictions:
Next, we need to identify any [tex]\( x \)[/tex]-values that make the denominator zero because these values will indicate vertical asymptotes and are not in the domain of the function:
[tex]\[
x^2(x + 3)(x + 1) = 0
\][/tex]
This equation simplifies to the factors:
[tex]\[
x^2 = 0 \quad \text{or} \quad x + 3 = 0 \quad \text{or} \quad x + 1 = 0
\][/tex]
Which gives us:
[tex]\[
x = 0 \quad \text{or} \quad x = -3 \quad \text{or} \quad x = -1
\][/tex]
Therefore, there are vertical asymptotes at [tex]\( x = 0 \)[/tex], [tex]\( x = -3 \)[/tex], and [tex]\( x = -1 \)[/tex].
3. Validating the Zeros:
Since zeros of the function cannot coincide with the vertical asymptotes, we disregard [tex]\( x = 0 \)[/tex] because it makes the denominator zero. Hence, it is not a valid zero.
The only valid zero that remains is [tex]\( x = 1 \)[/tex].
4. Determining the [tex]\( x \)[/tex]-Intercepts:
The [tex]\( x \)[/tex]-intercepts of the function are the points where the graph crosses the [tex]\( x \)[/tex]-axis, which occur at the valid zeros of the function. For [tex]\( f(x) = \frac{3x(x - 1)}{x^2(x + 3)(x + 1)} \)[/tex], the only valid zero is [tex]\( x = 1 \)[/tex].
Therefore, the function has an [tex]\( x \)[/tex]-intercept at:
[tex]\[
(1, 0)
\][/tex]
5. Summary:
- The real zero of the function is [tex]\( x = 1 \)[/tex].
- The corresponding [tex]\( x \)[/tex]-intercept is [tex]\( (1,0) \)[/tex].
- The vertical asymptotes of the function are at [tex]\( x = -3 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = 0 \)[/tex].
- There is no [tex]\( x \)[/tex]-intercept at [tex]\( x = 0 \)[/tex] or [tex]\( x = -3 \)[/tex].
Hence, the graph of the function [tex]\( f(x) = \frac{3x(x-1)}{x^2(x+3)(x+1)} \)[/tex] has one [tex]\( x \)[/tex]-intercept at [tex]\( (1,0) \)[/tex] and vertical asymptotes at [tex]\( x = -3 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = 0 \)[/tex].
