Answer :
Answer with Step-by-step explanation:
We are given that a function
[tex]f(x)=\frac{-2x+5}{x^2-5x-6}[/tex]
1.[tex]f(x)=\frac{-2x+5}{(x-6)(x+1)}[/tex]
The given function is not define at x=-1 and x=6.
Therefore, at x=-1 and at x=6 limit of given function does not exist.
2.Domain of f=R-{-1,6}
3.[tex]\lim_{x\rightarrow \infty}f(x)=\lim_{x\rightarrow \infty}\frac{x(-2+\frac{5}{x})}{x^2(1-\frac{5}{x}-\frac{6}{x^2})}=0[/tex]
When x approaches infinity then function approach to zero.
4.Vertical asymptote :
Substitute denominator =0
[tex](x-6)(x+1)=0[/tex]
[tex]x+1=0\implies x=-1[/tex]
[tex]x-6=0\implies x=6[/tex]
Horizontal asymptote:
The degree of numerator=1
Degree of denominator=2
Degree of numerator is less than the degree of denominator.Therefore,
Horizontal asymptote=0
Answer:
1. It is a rational function.
2. [tex]Domain=(-\infty,-1)\cup (-1,6)\cup (6,\infty)[/tex].
3. End behaviors are shown below.
4. Vertical asymptotes : x=6 and x=-1; horizontal asymptote : y=0.
Step-by-step explanation:
The given function is
[tex]f(x)=\dfrac{-2x+5}{x^2-5x-6}[/tex]
1.
A ration function is defined as
f(x)=p(x)/q(x)
The given function is in the form of p(x)/q(x).
Therefore, it is a rational function.
Find the factors of denominator.
[tex]f(x)=\dfrac{-2x+5}{x^2-6x+x-6}[/tex]
[tex]f(x)=\dfrac{-2x+5}{x(x-6)+1(x-6)}[/tex]
[tex]f(x)=\dfrac{-2x+5}{(x-6)(x+1)}[/tex]
Equate the denominator equal to 0.
[tex](x-6)(x+1)=0[/tex]
[tex]x=6,-1[/tex]
Therefore, the function is not defined for x=6 and x=-1.
2.
The domain of given function is all real number except 6 and -1.
[tex]Domain=(-\infty,-1)\cup (-1,6)\cup (6,\infty)[/tex]
3.
End behavior of the function.
[tex]f(x)\rightarrow 0\text{ as } x\rightarrow -\infty[/tex]
[tex]f(x)\rightarrow \infty\text{ as } x\rightarrow -1^-[/tex]
[tex]f(x)\rightarrow -\infty\text{ as } x\rightarrow -1^+[/tex]
[tex]f(x)\rightarrow \infty\text{ as } x\rightarrow 6^-[/tex]
[tex]f(x)\rightarrow -\infty\text{ as } x\rightarrow 6^+[/tex]
[tex]f(x)\rightarrow 0\text{ as } x\rightarrow \infty[/tex]
4.
To find the vertical asymptotes we need to find the zeroes of the denominator.
The vertical asymptotes of the function are x=6 and x=-1.
If degree of denominator is more than degree of numerator, then the horizontal asymptote is y=0.
