1) y ={ 1/x+1 } +4 2) 2x+8/3x-9 Domain Range V Asym H Asym Y-Intercept Graph:
I need help here @goformit100
Madam, sorry but can you wait for me, I am solving it.
Sure no problem
Madam See It may help you out http://www.google.co.in/#output=search&sclient=psy-ab&q=%282x%2B8%29%283x%2B9%29&oq=2x%2B8%2F3x-9&gs_l=serp.1.0.0i30j0i8i30.10962.10962.1.15057.1.1.0.0.0.0.178.178.0j1.1.0...0.0...1c.1.9.psy-ab.2OscIoceJ04&pbx=1&bav=on.2,or.r_qf.&fp=b3dafbcafab265ad&biw=1024&bih=629
Is that for the first 1? And I need to fill out the domain & range below .
It was for the second one.
Okay thank you. Domain Range V Asym H Asym Y-Intercept
Asym will be denoting the line satisfying and touching the Maximum point of the graph
There are two possibilities because you didn't really specify the problem. Either Y = 1/(X+1) OR Y = 1/X + 1 Either way, firs tyou substitute 4 in for Y in the first equation. For the first possibility 4 = 1/(X+1) 4(X+1) = 1 (Multiply both sides by X+1) 4X + 4 = 1 (Distribute) 4X = -3 (Subtract 4) X = -3/4 (Divide by 4) Second possibility 4 = 1/X +1 3 = 1/X (Subtract 1) 3X = 1 (Multiply by X) X = 1/3 (Divide by 3)
For number one right?
Oh it's not solving for x . It's graphing the equation. @steffanie1995
ohh my bad.. Hold up
@steffanie1995 yes madam plz go through the question again.
Sorry I don't know. I'm bad at math.
Haha no problem as well am I.
Domain for both the equations will be Real Numbers
Did you understood till here. ?
Hmmm a little bit.
And what left to be unanswered Plz Mention it madam.
1. Domain: What value makes denominator zero? Solve x + 1 = 0 That value must be excluded from domain.
For the second graph I have figured out: D: \[x \neq \frac{ 2 }{ 3 }\] R: \[y \neq 3\] V Asym: 3 H Asym: 2/3 Y Intercept : ?
Your above solutions are correct
Now its the matter of Y...
@mathstudent55 Is the above correct?
Yes above one is correct
Hey its correct, no worries @Ozzy733
For y-intercept just think of x=0, so \[y=\frac{2x+8}{3x-9} \\ \\ y=\frac{2(0)+8}{3(0)-9} \\ \\ y=-\frac{8}{9}\] \[\large (0,-\frac{8}{9})\]
Thanks! While I solve the y-intercept for the first graph, could you tell me what the h asym and domain would be for the first graph? As well as the graph sketch.
Would the y-intercept for the first graph just be 5?
@.Sam. ^
For the first one is it\[y=\frac{1}{x+1}+4\]
@Ozzy733 @.Sam.
Yes it is
It's a reciprocal with the graph shifted upwards by 4
yes the y-intercept is 5
Okay great! I'm stuck on the domain & hasym. I know it's a/c but I don't understand how to get it from this setup.
For Horizontal asymptote, as "x" approaches -1, the equation becomes infinity For Vertical asymptote, as "x" approaches infinity, the \(\huge\frac{1}{x+1}\) becomes too small and we can just call it zero, so it leaves with 4.
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So the asym is 0?
Thanks so much!
H:-1 V:4
welcome :)
So then the domain would be \[x \neq 4\] correct?
Yes the domain would be x≠4 is correct.
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