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f(x) = (x-5)/(2x^(2)-5x-3 )

Domain:
V.A:
Roots:
Y-int:
H.A:
Holes:
O.A:

Also, draw on the graph.

f(x) = (x-5)/(2x^(2)-5x-3 ) Domain: V.A: Roots: Y-int: H.A: Holes: O.A: Also, draw-example-1

2 Answers

1 vote

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.

Explanation:

We have given the function:


(x-5)/(2x^(2)-5x-3 )

So, first we simplify the equation we get,


(x-5)/((x-3)(2x+1))

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 :


\lim_(x \to \infty) (x-5)/(2x^(2)-5x-3)=0

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.

f(x) = (x-5)/(2x^(2)-5x-3 ) Domain: V.A: Roots: Y-int: H.A: Holes: O.A: Also, draw-example-1
User Ferhan
by
8.3k points
1 vote

i) The given function is


f(x)=(x-5)/(2x^2-5x-3)

The factored form is


f(x)=(x-5)/((x-3)(2x+1))

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


(x-3)(2x+1)\\e 0


(x-3)\\e0,(2x+1)\\e 0


x\\e3,x\\e-(1)/(2)

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


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


(x-3)=\\e0,(2x+1)=0


x=3,x=-(1)/(2)

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


x-5=0

The root is
x=5

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


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


f(0)=(-5)/((-3)(1))


f(0)=(5)/(3)

The y-intercept is
(5)/(3)

v) The horizontal asymptote is given by;


lim_(x\to \infty)(x-5)/(2x^2-5x-3)=0

The horizontal asymptote is
y=0

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.

f(x) = (x-5)/(2x^(2)-5x-3 ) Domain: V.A: Roots: Y-int: H.A: Holes: O.A: Also, draw-example-1
User James Martineau
by
8.7k points

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