Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
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].
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].
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.