Question

how would you describe the relationship between the real zero(s) and x-intercept(s) of the function \\( r(x) = \frac{3x(x - 1)}{x^3(x + 3)(x + 1)} \\)\
\\( \bigcirc \\) when you set the function equal to zero, the solution is \\( x = 1 \\); therefore, the graph has an x-intercept of \\( (1, 0) \\).\
\\( \bigcirc \\) when you set the function equal to zero, the solutions are \\( x = 0 \\) or \\( x = 1 \\); therefore, the graph has x-intercepts at \\( (0, 0) \\) and \\( (1, 0) \\).\
\\( \bigcirc \\) when you substitute \\( x = 0 \\) into the function, there is no solution; therefore, the graph will not have any x-intercepts.\
\\( \bigcirc \\) since there are asymptotes at \\( x = -3 \\), \\( x = -1 \\), and \\( x = 0 \\), the graph has no x-intercepts and, therefore, no real zeros.

Explanation:

Step1: Find real zeros

To find real zeros, set \( R(x) = 0 \). The function is \( R(x)=\frac{3x(x - 1)}{x^{2}(x + 3)(x + 1)} \). A fraction is zero when numerator is zero (denominator non - zero). So solve \( 3x(x - 1)=0 \).
The solutions are \( x = 0 \) or \( x=1 \). But we need to check the denominator. When \( x = 0 \), the denominator \( x^{2}(x + 3)(x + 1)=0 \), so \( x = 0 \) makes the function undefined. So we discard \( x = 0 \). The only valid real zero is \( x = 1 \).

Step2: Find x - intercepts

The x - intercept of a function \( y = f(x) \) is the point where \( y = 0 \), i.e., \( (x,0) \) where \( x \) is the real zero of the function. Since the real zero is \( x = 1 \), the x - intercept is \( (1,0) \).
Now let's analyze the options:

• Option 1: When we set the function equal to zero, we solve \( \frac{3x(x - 1)}{x^{2}(x + 3)(x + 1)}=0 \). Numerator \( 3x(x - 1)=0\Rightarrow x = 0 \) or \( x = 1 \). But \( x = 0 \) makes denominator zero, so only \( x = 1 \) is valid. So the x - intercept is \( (1,0) \). This option is correct.
• Option 2: \( x = 0 \) is not a valid zero because it makes the denominator zero, so we can't have an x - intercept at \( (0,0) \).
• Option 3: Substituting \( x = 0 \) into the function gives an undefined value, but we find zeros by setting the function equal to zero, not by substituting \( x = 0 \) and checking for a solution in that way. The function can have a zero at a point where it is defined, and we found \( x = 1 \) is a zero.
• Option 4: Asymptotes are related to where the function is undefined (denominator zero), but the numerator can be zero at a point where the function is defined (after excluding points that make denominator zero). We found \( x = 1 \) is a zero, so the function has an x - intercept.

Answer:

When you set the function equal to zero, the solution is \( x = 1 \); therefore, the graph has an x - intercept of \( (1,0) \) (the first option).