For each,identify the ×- and y- intercepts the asymptotes,and the intervals where the function is above o below the ×-axis.Sketch the graph.Idebtify the domain.
A) f(×)=(5×-2)(×-2)/(3×-4)(×+2)
B) f(×)=(ײ-+6)/(ײ-6×+8)
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Answers
Answer:
SOLUTION
First, factor the numerator and denominator.
⎧
⎪
⎨
⎪
⎩
k
(
x
)
=
5
+
2
x
2
2
−
x
−
x
2
=
5
+
2
x
2
(
2
+
x
)
(
1
−
x
)
To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:
{
(
2
+
x
)
(
1
−
x
)
=
0
x
=
−
2
,
1
Neither \displaystyle x=-2x=−2 nor \displaystyle x=1x=1 are zeros of the numerator, so the two values indicate two vertical asymptotes. Figure 9 confirms the location of the two vertical asymptotes.
Graph of k(x)=(5+2x)^2/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.
Figure 9
Removable Discontinuities
Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity.
For example, the function \displaystyle f\left(x\right)=\frac{{x}^{2}-1}{{x}^{2}-2x - 3}f(x)=
x
2
−2x−3
x
2
−1
may be re-written by factoring the numerator and the denominator.
\displaystyle f\left(x\right)=\frac{\left(x+1\right)\left(x - 1\right)}{\left(x+1\right)\left(x - 3\right)}f(x)=
(x+1)(x−3)
(x+1)(x−1)
Notice that \displaystyle x+1x+1 is a common factor to the numerator and the denominator. The zero of this factor, \displaystyle x=-1x=−1, is the location of the removable discontinuity. Notice also that \displaystyle x - 3x−3 is not a factor in both the numerator and denominator. The zero of this factor, \displaystyle x=3x=3, is the vertical asymptote.
Graph of f(x)=(x^2-1)/(x^2-2x-3) with its vertical asymptote at x=3 and a removable discontinuity at x=-1.