Part D
we have the function
[tex]\begin{gathered} y=\frac{4}{-(x+1)}+3=\frac{4-3(x+1)}{-(x+1)}=\frac{4-3x-3}{-(x+1)}=\frac{1-3x}{-(x+1)}=\frac{3x-1}{x+1} \\ \\ y=\frac{3x-1}{x+1} \end{gathered}[/tex]
In this rational function
Remember that
The denominator cannot be equal to zero
so
The value of x cannot be equal to x=-1
At x=-1 there is a vertical asymptote
Find out a horizontal asymptote
Degree on Top is Equal to the Bottom
so
the horizontal asymptote is at y=3/1=3
Find out the intercepts
y-intercept (value of y when the value of x=0)
For x=0
[tex]y=\frac{3(0)-1}{0+1}=-1[/tex]
The y-intercept is (0,-1)
Find out the x-intercept (value of x when the value of y=0)
For y=0
[tex]\begin{gathered} 0=\frac{3x-1}{x+1} \\ \\ 3x-1=0 \\ 3x=1 \\ x=\frac{1}{3} \end{gathered}[/tex]
The x-intercept is (0.33,0)
With the given information
Graph the function
using a graphing tool
see the figure below
