Question

Consider the function f(x) = −2x + 5 / x^2 − 5x − 6.

1. Looking at the structure of the function, what information can you gather about the graph of f?
2. State the domain of f
3. Determine the end behavior of f.
4. State the equations of any vertical and horizontal asymptotes on the graph of y = f(x).

Answer 1

1. It is a rational function.

2.
`Domain=(-\infty,-1)\cup (-1,6)\cup (6,\infty)`.

3. End behaviors are shown below.

4. Vertical asymptotes : x=6 and x=-1; horizontal asymptote : y=0.

Explanation:

The given function is

`f(x)=(-2x+5)/(x^2-5x-6)`

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.

`f(x)=(-2x+5)/(x^2-6x+x-6)`

`f(x)=(-2x+5)/(x(x-6)+1(x-6))`

`f(x)=(-2x+5)/((x-6)(x+1))`

Equate the denominator equal to 0.

`(x-6)(x+1)=0`

`x=6,-1`

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.

`Domain=(-\infty,-1)\cup (-1,6)\cup (6,\infty)`

3.

End behavior of the function.

`f(x)\rightarrow 0\text{ as } x\rightarrow -\infty`

`f(x)\rightarrow \infty\text{ as } x\rightarrow -1^-`

`f(x)\rightarrow -\infty\text{ as } x\rightarrow -1^+`

`f(x)\rightarrow \infty\text{ as } x\rightarrow 6^-`

`f(x)\rightarrow -\infty\text{ as } x\rightarrow 6^+`

`f(x)\rightarrow 0\text{ as } x\rightarrow \infty`

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.

Answer 2

Answer with Step-by-step explanation:

We are given that a function

`f(x)=(-2x+5)/(x^2-5x-6)`

1.
`f(x)=(-2x+5)/((x-6)(x+1))`

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.
`\lim_(x\rightarrow \infty)f(x)=\lim_(x\rightarrow \infty)(x(-2+(5)/(x)))/(x^2(1-(5)/(x)-(6)/(x^2)))=0`

When x approaches infinity then function approach to zero.

4.Vertical asymptote :

Substitute denominator =0

`(x-6)(x+1)=0`

`x+1=0\implies x=-1`

`x-6=0\implies x=6`

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