Given:
The function is:
[tex]f(x)=\dfrac{1}{x+3}-2[/tex]
To find:
The graph of the given function.
Solution:
We have,
[tex]f(x)=\dfrac{1}{x+3}-2[/tex]
It can be written as:
[tex]f(x)=\dfrac{1-2(x+3)}{x+3}[/tex]
[tex]f(x)=\dfrac{1-2x-6}{x+3}[/tex]
[tex]f(x)=\dfrac{-2x-5}{x+3}[/tex]
Putting [tex]x=0[/tex] to find the y-intercept.
[tex]f(0)=\dfrac{-2(0)-5}{(0)+3}[/tex]
[tex]f(0)=\dfrac{-5}{3}[/tex]
So, the y-intercept is [tex]\dfrac{-5}{3}[/tex].
Putting [tex]f(x)=0[/tex] to find the x-intercept.
[tex]0=\dfrac{-2x-5}{x+3}[/tex]
[tex]0=-2x-5[/tex]
[tex]2x=-5[/tex]
[tex]x=\dfrac{-5}{2}[/tex]
[tex]x=-2.5[/tex]
So, the x-intercept is [tex]-2.5[/tex].
For vertical asymptote, equate the denominator and 0.
[tex]x+3=0[/tex]
[tex]x=-3[/tex]
So, the vertical asymptote is [tex]x=-3[/tex].
The degrees of numerator and denominator are equal, so the horizontal asymptote is the ratio of leading coefficients.
[tex]y=\dfrac{-2}{1}[/tex]
[tex]y=-2[/tex]
So, the horizontal asymptote is [tex]y=-2[/tex].
End behavior of the given function:
[tex]f(x)\to -2[/tex] as [tex]x\to -\infty[/tex]
[tex]f(x)\to -\infty[/tex] as [tex]x\to -3^-[/tex]
[tex]f(x)\to \infty[/tex] as [tex]x\to -3^+[/tex]
[tex]f(x)\to -2[/tex] as [tex]x\to \infty[/tex]
Using all these key features, draw the graph of given function as shown below.
