Respuesta :

kudzordzifrancis

i) The given function is

[tex]f(x)=\frac{x-5}{2x^2-5x-3}[/tex]

The factored form is

[tex]f(x)=\frac{x-5}{(x-3)(2x+1)}[/tex]

The domain are the values of  x for which the function is defined.

[tex](x-3)(2x+1)\ne 0[/tex]

[tex](x-3)\ne0,(2x+1)\ne 0[/tex]

[tex]x\ne3,x\ne-\frac{1}{2}[/tex]

ii) To find the vertical asymptotes, equate the denominator to zero.

[tex](x-3)(2x+1)=0[/tex]

[tex](x-3)=\ne0,(2x+1)=0[/tex]

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

iii) To find the roots, equate the numerator to zero.

[tex]x-5=0[/tex]

The root is [tex]x=5[/tex]

iv) To find the y-intercept, put [tex]x=0[/tex] into the function.

[tex]f(0)=\frac{0-5}{(0-3)(2(0)+1)}[/tex]

[tex]f(0)=\frac{-5}{(-3)(1)}[/tex]

[tex]f(0)=\frac{5}{3}[/tex]

The y-intercept is [tex]\frac{5}{3}[/tex]

v) The horizontal asymptote is given by;

[tex]lim_{x\to \infty}\frac{x-5}{2x^2-5x-3}=0[/tex]

The horizontal asymptote is [tex]y=0[/tex]

vi) The function is not reducible. There are no holes.

vii) The given function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

Ver imagen kudzordzifrancis
zainebamir540

Answer:

Domain:All real numbers except  x≠3 and  x≠ -1/2,V.A= x=3 , x= -1/2.root=5,y-int= 5/3.H.A=0 , no hole , no  O.A.

Step-by-step explanation:

We  have given the function:

[tex]\frac{x-5}{2x^{2}-5x-3 }[/tex]

So, first we simplify the equation we get,

[tex]\frac{x-5}{(x-3)(2x+1)}[/tex]

We have to find the domain :

Domain is the value of x for which the function is defined.

For this, denominator must not be equal to zero.

(x-3)(2x+1) ≠ 0

(x-3)≠0 , (2x+1)≠0

x≠3 , x≠ -1/2

So, the domain is all real numbers except  x≠3 and  x≠ -1/2.

For vertical asymptotes, denominator is equal to zero.

(x-3)(2x+1) = 0

x=3 , x= -1/2 are  vertical asymptotes.

Nominator is equal to zero  for roots:

x-5 = 0

x= 5 is the root.

Put x = 0 for y-intercept:

y = 0-5/(0-3)(2(0)+1)

y = 5/3 is the y-intercept.

Horizontal asymptote is :

[tex]\lim_{x \to \infty} \frac{x-5}{2x^{2}-5x-3}=0[/tex]

Horizontal asymptote is  y = 0.

The function has no hole because it is reducible.

It has no oblique asymptote because it is proper function.

Graph is attached.

Ver imagen zainebamir540

Otras preguntas