Question

3 (lt 2.4) which figure below is a graph of ( r(x) = 2 + \frac{x + 1}{x - 3} )? show your thinking!

a.
graph a

b.
graph b with +0.5 annotation

c.
graph c with 2/4 annotation

d.
graph d

explain how you know. make sure to describe both vertical and horizontal asymptotes of ( r(x) ).
i chose b because the vertical asymptote is ( x = 3 ) which means ( x ) cannot be 3 or it would be undefined and i found the y - intercept where the graph crosses.

4 (lt 2.5) solve the equations below for ( x ). show all your work!

a. ( \frac{x + 5}{x - 1} = 4(x - 1) )
work shown: ( x + 5 = 4x - 4 ), etc. with check marks and 2/2 annotation

b. ( \frac{3}{x + 5} = \frac{3}{x(x + 5)} )
work shown with its just x annotation and other steps

Explanation:

Step1: Identify Vertical Asymptote

For the function \( r(x) = 2+\frac{x + 1}{x - 3}\), the vertical asymptote occurs where the denominator is zero (and numerator is non - zero). Set \(x-3 = 0\), so \(x = 3\). This means the vertical asymptote is \(x = 3\). We look for the graph where there is a vertical dashed line at \(x = 3\).

Step2: Identify Horizontal Asymptote

To find the horizontal asymptote of a rational function \(y=\frac{f(x)}{g(x)}\) (after simplifying the given function), we compare the degrees of the numerator and denominator. First, simplify \(r(x)=2+\frac{x + 1}{x - 3}=\frac{2(x - 3)+x + 1}{x - 3}=\frac{2x-6+x + 1}{x - 3}=\frac{3x-5}{x - 3}\). The degree of the numerator (\(3x-5\)) and the denominator (\(x - 3\)) is 1 (since the highest power of \(x\) in both is 1). When the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. The leading coefficient of the numerator is 3 and of the denominator is 1, but wait, let's re - express the original function in another way. \(r(x)=2+\frac{x + 1}{x - 3}\), as \(x\to\pm\infty\), \(\frac{x + 1}{x - 3}\to\frac{x}{x}=1\) (dividing numerator and denominator by \(x\) for large \(x\)). So \(r(x)\to2 + 1=3\) as \(x\to\pm\infty\). So the horizontal asymptote is \(y = 3\)? Wait, no, wait the original simplification: \(r(x)=2+\frac{x + 1}{x - 3}=\frac{2(x - 3)+x + 1}{x - 3}=\frac{2x-6+x + 1}{x - 3}=\frac{3x - 5}{x - 3}\). Divide numerator and denominator by \(x\): \(\frac{3-\frac{5}{x}}{1-\frac{3}{x}}\). As \(x\to\pm\infty\), \(\frac{5}{x}\to0\) and \(\frac{3}{x}\to0\), so \(r(x)\to\frac{3}{1}=3\)? Wait, but let's check the initial form. \(r(x)=2+\frac{x + 1}{x - 3}\), \(\frac{x + 1}{x - 3}=\frac{x-3+4}{x - 3}=1+\frac{4}{x - 3}\), so \(r(x)=2 + 1+\frac{4}{x - 3}=3+\frac{4}{x - 3}\). Ah, that's a better simplification. So as \(x\to\pm\infty\), \(\frac{4}{x - 3}\to0\), so \(r(x)\to3\). So horizontal asymptote is \(y = 3\). Now, looking at the graphs, graph b (the middle - right graph) has a vertical asymptote at \(x = 3\) (dashed line at \(x = 3\)) and the horizontal asymptote should be \(y = 3\) (the horizontal line that the graph approaches). Also, we can check the \(y\) - intercept: when \(x = 0\), \(r(0)=2+\frac{0 + 1}{0 - 3}=2-\frac{1}{3}=\frac{5}{3}\approx1.67\). Graph b also matches the \(y\) - intercept behavior.

Step3: Match the Graph

Among the given graphs, graph b (the one in the top - right with the dashed line at \(x = 3\)) has a vertical asymptote at \(x = 3\) (since the denominator is zero at \(x = 3\)) and as \(x\) approaches \(\pm\infty\), the function approaches \(y = 3\) (horizontal asymptote). Also, the shape of the hyperbola - like graph (since it's a rational function) with two branches, one on the left of \(x = 3\) and one on the right, matches the form of \(r(x)=3+\frac{4}{x - 3}\) (which has a vertical asymptote at \(x = 3\) and horizontal at \(y = 3\)).

Answer:

The graph labeled 'b' (the top - right graph) is the graph of \(r(x)=2+\frac{x + 1}{x - 3}\)