This Post Has 8 Comments

1. Number 3, because if you replace x for 1 then you get zero. and because it is rise over run then the run is 0. it is the only line with a run of 0

   Reply

2. The graph as shown below.

   Step-by-step explanation:

   Given : The function [tex]f(x)=\frac{1}{x-3}-2[/tex]

   We have to plot the graph for the given function   [tex]f(x)=\frac{1}{x-3}-2[/tex]  

   Consider the given function [tex]f(x)=\frac{1}{x-3}-2[/tex]

   Domain of function [tex]f(x)=\frac{1}{x-3}-2[/tex]

   DOMAIN is set of input values for which the function is real and has defined values.        

   So, The given function is undefined at x = 3

   So, Domain is [tex]x<3\quad \mathrm{or}\quad \:x3[/tex]

   RANGE is the set of values of dependent variable  for which the function is defined.

   Inverse of given function is  [tex]y=\frac{3x+7}{x+2}[/tex]

   Now, domain of inverse function is [tex]f\left(x\right)<-2\quad \mathrm{or}\quad \:f\left(x\right)-2[/tex]

   Now, x intercept and y- intercepts

   x intercept where y = 0 and y- intercept where x= 0

   Let f(x) = y

   Then [tex]y=\frac{1}{x-3}-2[/tex]

   Put x = 0

   thus y- intercept is [tex]\left(0,\:-\frac{7}{3}\right)[/tex]

   Now put y = 0

   Then  x- intercept is [tex]\left(\frac{7}{2},\:0\right)[/tex]

   Now, Calculate the vertical and horizontal asymptotes,

   Vertical asymptotes,

   Go over every undefined point and check if at least one of the following statements is satisfied.

   [tex]\lim _{x\to a^-}f\left(x\right)=\pm \infty[/tex]

   [tex]\lim _{x\to a^+}f\left(x\right)=\pm \infty[/tex]

   Thus, The vertical asymptotes is x = 3

   And For horizontal asymptotes,

   [tex]\mathrm{Check\:if\:at\:}x\to \pm \infty \mathrm{\:the\:function\:}y=\frac{1}{x-3}-2\mathrm{\:behaves\:as\:a\:line,\:}y=mx+b[/tex]

   We have y = -2 as horizontal asymptotes.

   Plot we get the graph as shown below.

   [tex]Which graph represents the function f(x)=1/x+3-2[/tex]

   Reply

3. files are attached below

   [tex]Which graph represents the function f(x)=1/x+3-2[/tex]
   [tex]Which graph represents the function f(x)=1/x+3-2[/tex]

   Reply

4. To quickly solve this problem, we can use a graphing tool or a calculator to plot the equation.

   Please see the attached image below, to find more information about the graph

   The equation is:

   f(x) = 1/x - 1

   Domain

   All real numbers except for {0}

   [tex]Which graph represents the function f(x) = 1/x - 1?[/tex]

   Reply

5. It would be the fourth graph, D

   Explanation:

   When faced with these problems, you can either plot it on your own or you can use a graph generator online.

   Reply

6. Step-by-step explanation:

   To eliminate ambiguity please use parentheses when typing a function like this one.  I'm going to assume that you meant

   f(x) = 1 / (x + 3)     -      2

   Start with the graph of y = 1/x.  Its vertical asymptote is the line x = 0.  The graph never touches the x-axis.  

   Translating the entire graph 3 units to the left will give you the function

   h(x) = 1 / (x + 3).

   Translating the above graph 2 units down will give you the desired function g(x) = 1 / (x + 3) - 2.

   Reply

7. Plot the graph: [tex]y=\dfrac{1}{x}[/tex]

   Shift it 3 units left and 2 units down.

   f(x + n) - shift n units left

   f(x - n) - shift n units right

   f(x) + n - shift n units up

   f(x) - n - shift n units down

   [tex]What graph represents the function f(x)=1/x+3-2[/tex]

   Reply

8. Figure 2

   Step-by-step explanation:

   We are given the function [tex]f(x)=\frac{1}{x-1}[/tex].

   Now, when x = 1, we have that,

   [tex]f(x)=\frac{1}{1-1}[/tex] i.e. [tex]f(x)=\frac{1}{0}[/tex] i.e. [tex]f(x)\rightarrow\infty[/tex].

   Also, when x = -1, we have that,

   [tex]f(x)=\frac{1}{-1-1}[/tex] i.e. [tex]f(x)=\frac{1}{-2}[/tex].

   Further, when f(x) = -1, we have that,

   [tex]-1=\frac{1}{x-1}[/tex] i.e. [tex]-x+1=1[/tex] i.e. [tex]x=0[/tex].

   So, it is clear from the given options that the second figure is the correct graph of the function [tex]f(x)=\frac{1}{x-1}[/tex].

   Hence, figure 2 is the correct option.

   Reply