Question

given the functions f(x) = x² - 4 and g(x) = x - 1, a graph of the combined function y = (f(x))/(g(x)) most likely resembles.

Explanation:

Step1: Simplify the combined function

First, substitute \( f(x) = x^2 - 4 \) and \( g(x)=x - 1 \) into \( y=\frac{f(x)}{g(x)} \). We know that \( x^2 - 4 \) can be factored as a difference of squares: \( x^2 - 4=(x - 2)(x + 2) \). So the function becomes \( y=\frac{(x - 2)(x + 2)}{x - 1} \), with the domain restriction \( x
eq1 \) (since division by zero is undefined).

Step2: Analyze the function's behavior

• Vertical Asymptote: Since the denominator is zero when \( x = 1 \) and the numerator is not zero at \( x = 1 \) (substituting \( x = 1 \) into numerator: \( (1 - 2)(1 + 2)=(-1)(3)=-3

eq0 \)), there is a vertical asymptote at \( x = 1 \).

• x - intercepts: Set the numerator equal to zero: \( (x - 2)(x + 2)=0 \), so \( x = 2 \) or \( x=-2 \). These are the x - intercepts (points \( (-2,0) \) and \( (2,0) \)).
• Behavior near vertical asymptote:
• For \( x\to1^+ \), the denominator \( x - 1\to0^+ \), and the numerator \( (x - 2)(x + 2)\to(1 - 2)(1 + 2)=-3 \), so \( y\to-\infty \).
• For \( x\to1^- \), the denominator \( x - 1\to0^- \), and the numerator \( (x - 2)(x + 2)\to - 3 \), so \( y\to+\infty \).
• End - behavior: We can perform polynomial long - division or rewrite the function. Let's rewrite \( y=\frac{x^2-4}{x - 1}=\frac{x^2 - x+x - 1-3}{x - 1}=\frac{x(x - 1)+(x - 1)-3}{x - 1}=x + 1-\frac{3}{x - 1} \). As \( x\to\pm\infty \), the term \( \frac{3}{x - 1}\to0 \), so the function behaves like the line \( y=x + 1 \) for large \( |x| \).

To identify the graph, we look for a graph with a vertical asymptote at \( x = 1 \), x - intercepts at \( x=-2 \) and \( x = 2 \), and end - behavior similar to \( y=x + 1 \), with the appropriate behavior near \( x = 1 \).

(Note: Since the problem is about identifying the graph of a rational function, we would compare the above - derived features with the given graphs. For example, if there are three graphs: one with vertical asymptote at \( x = 1 \), x - intercepts at \( x=-2 \) and \( x = 2 \), and slant - like end - behavior, that would be the correct one. If we assume the graphs are labeled as Left, Middle, Right, and the Middle graph has these features, the answer would be the Middle Graph.)

Answer:

The Middle Graph (assuming the graph with vertical asymptote \( x = 1 \), x - intercepts at \( x=-2,x = 2 \), and end - behavior like \( y=x + 1 \) is the middle one among the given options)