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1 vote

f(x) = (2x-1)/(x^(2)-x-6 )

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

Also draw on the graph.

f(x) = (2x-1)/(x^(2)-x-6 ) Domain: V.A: Roots: Y-Int: H.A: Holes: O.A: Also draw on-example-1
User Semyazas
by
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1 Answer

6 votes

i) The given function is


f(x)=(2x-1)/(x^2-x-6)

We factor to obtain


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

The domain is


(x-3)(x+2)\\e0


(x-3)\\e0,(x+2)\\e0


x\\e3,x\\e-2

ii) The vertical asymptotes are


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


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


x=3,x=-2

iii) To find the root, we equate the numerator to zero.


2x-1=0


x=(1)/(2)

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


f(0)=(2(0)-1)/((0)^2-(0)-6)


f(0)=(-1)/(-6)


f(0)=(1)/(6)

vi) To find the horizontal asymptote, we take limit to infinity.

This implies that;


lim_(x\to \infty)(2x-1)/(x^2-x-6)=0

The horizontal asymptote is y=0.

vii) The numerator and the denominator do not have common factors that are at least linear.

Therefore the function has no holes in it.

f(x) = (2x-1)/(x^(2)-x-6 ) Domain: V.A: Roots: Y-Int: H.A: Holes: O.A: Also draw on-example-1
User Shontelle
by
8.5k points

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