How would you describe the relationship between the real zero(s) and the [tex]$x$[/tex]-intercept(s) of the function

[tex]\[ f(x) = \frac{3x(x-1)}{x^2(x+3)(x+1)} \][/tex]

A. When you set the function equal to zero, the solution is [tex]$x = 1$[/tex]; therefore, the graph has an [tex]$x$[/tex]-intercept of [tex]$(1, 0)$[/tex].

B. When you set the function equal to zero, the solutions are [tex]$x = 0$[/tex] or [tex]$x = 1$[/tex]; therefore, the graph has [tex]$x$[/tex]-intercepts at [tex]$(0, 0)$[/tex] and [tex]$(1, 0)$[/tex].

C. When you substitute [tex]$x = 0$[/tex] into the function, there is no solution; therefore, the graph will not have any [tex]$x$[/tex]-intercepts.

D. Since there are asymptotes at [tex]$x = -3$[/tex], [tex]$x = -1$[/tex], and [tex]$x = 0$[/tex], the graph has no [tex]$x$[/tex]-intercepts and, therefore, no real zeros.



Answer :

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].