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# Rational function

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https://snag.gy/qfoDxR.jpg

i need help on question 3D please explain each steps

Sep 29, 2018

#1
+2

y  = [ x ( x - 2) ] / [ x^2 - 1 ] =    [x^2 - 2x] / [ (x + 1) ( x - 1) ]   = (x^2 - 2x) / [ x^2 - 1)

a)  Setting both linear factors in the denominator to 0 and solving for x gives us the vertical asyptotes of  x  = -1 and x  = 1

Because the coefficients on each of the x^2 terms in the numerator/ denominator  are 1, their ratio  gives the  horizontal asymptote of  y  = 1 / 1    =  1

b) To see where it crosses the  x axis, let y  = 0  and we have

0  = x (x - 2)  / (x^2 - 1)     multiply both sides by  x^2 - 1

0 =  x ( x - 2)   set both factors to  0  and solve for x and we get the x intercepts of x = 0  and  x  = 2

To see where it crosses the y  axis, let x  = 0   and we have  0 (0 - 2) / ( 0^2 - 1)   =  0

c) Here is the curve : https://www.desmos.com/calculator/i0vbehvizi

d) Inspection of the graph reveals that the intervals that solve this inequality  are  (-infinity ,-1) U [0, 1) U [2, infinity )   Sep 29, 2018
#2
+1

the final answer for D should be 𝑥 ≥ 2, 0 ≤ 𝑥 < 1, 𝑥 < −1

which is really confusing please explain

YEEEEEET  Sep 29, 2018
#3
+2

My notation means the same thing

x ≥ 2  is the same as  [2,  infinity )

x < -1   is the same  as   ( - infinity, -1)

0 ≤ 𝑥 < 1  is the same as  [0, -1 )   CPhill  Sep 29, 2018
#4
+1

sorry but can u please explain how u get those values as im really bad at graphs

YEEEEEET  Sep 29, 2018
#5
+2

OK...look at the graph....

Note that the graph is  above  y =  0  from  x  =   -infinity to x =  - 1   [but not including -1 ]

And the graph  is equal to  or above  y = 0  from x = 0   0 to x  =  1     [ icluding 0, but not including 1]

Ad the graph is equal to or above  y  = 0  from x =   2  to  x =  infinity  [  including 2 ]   CPhill  Sep 29, 2018
#6
+1

thank you so much! YEEEEEET  Sep 29, 2018