Question

essential question what are some of the characteristics of the grapt a rational function? the parent function for rational functions with a linear numerator and a linear denominator is f(x) = 1/x, parent function the graph of this function, shown at the right, is a hyperbola. 1 exploration: identifying graphs of rational functions work with a partner. each function is a transformation of the graph of the parent function f(x) = 1/x. match the function with its graph. explain your reasoning. ther describe the transformation. a. g(x) = 1/(x - 1) b. g(x) = -1/(x - 1) c. g(x) = . d. g(x) = (x - 2)/(x + 1) e. g(x) = x/(x + 2) f. g(x) = . a. graph a b. graph b

Explanation:

Step1: Analyze function a: \(g(x)=\frac{1}{x - 1}\)

The parent function is \(f(x)=\frac{1}{x}\). For \(g(x)=\frac{1}{x - 1}\), the transformation is a horizontal shift. The vertical asymptote of \(f(x)\) is \(x = 0\), for \(g(x)\) it is \(x=1\) (shift right 1 unit). The horizontal asymptote remains \(y = 0\). Looking at the graphs, we need to find a graph with vertical asymptote \(x = 1\) (but since the given graphs A and B have vertical asymptotes around -4 and others, maybe we re - evaluate. Wait, maybe the graphs are for other functions. Wait, let's take function b: \(g(x)=\frac{-1}{x - 1}\). This is a reflection over the x - axis and horizontal shift right 1. The vertical asymptote is \(x = 1\), horizontal asymptote \(y = 0\), and it's reflected.

Step2: Analyze graph A

Graph A has a vertical asymptote (dashed line) at \(x=-4\)? Wait, no, the dashed line in graph A is at \(x=-4\)? Wait, the x - axis has -6, -4, 6. Wait, maybe the vertical asymptote for graph A is \(x=-1\) (if we miscalculate). Wait, let's take function d: \(g(x)=\frac{x - 2}{x + 1}=\frac{(x + 1)-3}{x + 1}=1-\frac{3}{x + 1}\). The vertical asymptote is \(x=-1\), horizontal asymptote \(y = 1\). The graph of this function will have a vertical asymptote at \(x=-1\) and horizontal at \(y = 1\). Let's check graph A: it has a dashed vertical line (asymptote) and a horizontal dashed line (asymptote). If the horizontal asymptote is \(y = 1\) (close to the dashed line), and vertical at \(x=-1\) (the dashed vertical line). Function d: \(g(x)=\frac{x - 2}{x + 1}\), when \(x\to\pm\infty\), \(g(x)\to1\), so horizontal asymptote \(y = 1\), vertical asymptote \(x=-1\). Graph A has a horizontal asymptote (dashed line) and vertical asymptote (dashed line) at \(x=-1\) (the dashed vertical line is at \(x=-4\)? Wait, maybe I made a mistake. Let's take function a: \(g(x)=\frac{1}{x - 1}\) has vertical asymptote \(x = 1\), but the given graphs A and B have vertical asymptotes at negative x. Wait, maybe the graphs are for functions with vertical asymptotes at \(x=-1\) or \(x=-2\). Let's take function e: \(g(x)=\frac{x}{x + 2}=\frac{(x + 2)-2}{x + 2}=1-\frac{2}{x + 2}\), vertical asymptote \(x=-2\), horizontal asymptote \(y = 1\).

Wait, maybe the correct match for graph A: Let's look at the behavior. Graph A has two branches, one in the second quadrant (left of vertical asymptote) and one in the fourth quadrant (right of vertical asymptote) if the vertical asymptote is negative. Wait, function b: \(g(x)=\frac{-1}{x - 1}\) is a reflection of \(y=\frac{1}{x - 1}\) over the x - axis. So where \(y=\frac{1}{x - 1}\) is in the first and third quadrants (shifted right), \(y=\frac{-1}{x - 1}\) is in the second and fourth quadrants (shifted right). But the given graphs A and B: Graph A has a branch in the second quadrant (above x - axis, left of vertical asymptote) and a branch in the fourth quadrant (below x - axis? No, the graph in A has a branch in the second quadrant (increasing) and a branch in the first quadrant? Wait, no, the lower branch in graph A is in the fourth quadrant? Wait, maybe the correct match for graph A is function b: \(g(x)=\frac{-1}{x - 1}\) or function d.

Wait, let's start over. The parent function \(f(x)=\frac{1}{x}\) has vertical asymptote \(x = 0\), horizontal \(y = 0\), and is in first and third quadrants.

- For \(g(x)=\frac{1}{x - 1}\): shift right 1, vertical asymptote \(x = 1\), horizontal \(y = 0\), first and third quadrants (shifted).
- For \(g(x)=\frac{-1}{x - 1}\): reflect over x - axis, shift right 1, vertical asymptote \(x = 1\), horizontal \(y = 0\), second and fourth quadrants.
- For \(g(x)=\frac{x - 2}{x + 1}=1-\frac{3}{x + 1}\): vertical asymptote \(x=-1\), horizontal \(y = 1\), as \(x\to - 1^-\), \(g(x)\to+\infty\), as \(x\to - 1^+\), \(g(x)\to-\infty\), and as \(x\to\pm\infty\), \(g(x)\to1\).
- For \(g(x)=\frac{x}{x + 2}=1-\frac{2}{x + 2}\): vertical asymptote \(x=-2\), horizontal \(y = 1\), as \(x\to - 2^-\), \(g(x)\to+\infty\), as \(x\to - 2^+\), \(g(x)\to-\infty\), as \(x\to\pm\infty\), \(g(x)\to1\).

Looking at graph A: it has a vertical asymptote (dashed line) and a horizontal asymptote (dashed line). The horizontal asymptote is around \(y = 1\) (the dashed line), and vertical asymptote around \(x=-1\) (the dashed vertical line). So function d: \(g(x)=\frac{x - 2}{x + 1}\) has vertical asymptote \(x=-1\) and horizontal \(y = 1\), which matches graph A.

Answer:

If we are matching function d (\(g(x)=\frac{x - 2}{x + 1}\)) with graph A, then the answer is that function \(g(x)=\frac{x - 2}{x + 1}\) matches graph A. The reasoning is that \(g(x)=\frac{x - 2}{x + 1}=1-\frac{3}{x + 1}\), so it has a vertical asymptote at \(x=-1\) (the dashed vertical line in graph A) and a horizontal asymptote at \(y = 1\) (the dashed horizontal line in graph A). The transformation from \(f(x)=\frac{1}{x}\) is a horizontal shift left 1 unit, a vertical stretch by a factor of 3, and a vertical shift up 1 unit.

(Note: Since the problem was a bit unclear about which function - graph pair to match, we took a common example. If the intended function - graph pair was different, the reasoning would be adjusted accordingly.)