f(x)=2x2-x-3/x2-2x-3 graph and find the domain,range, and VA

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f(x)=2x2-x-3/x2-2x-3 graph and find the domain,range, and VA

Mathematics
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no hold on :)
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Do you mean \(\Large f(x)=\frac{2x^2−x−3}{x^2−2x−3}\)
yes:)
Do you know how to factorize quadratic expressions?
yes!
So I got for the first one, (x+1)(2x-3)
So factor the numerator and denominator, as a start! ok?
Got it:) So, I gottt (x+1)(2x-3)/(x+1)(x-3)
Yep! Good job! So you have: \(\Large f(x)=\frac{(2x-3)(x+1)}{(x-3)(x+1)}\) Do you notice anything that could help you?
Canceling out the x+1
Just like that, or with a condition?
I dont know what you mean by that? So, I'm going w/ "just like that."
Well, when you cancel factors in a division, you have to make sure that the factor will never become zero. So you can cancel ONLY if you specify x+1\(\ne\)0., in other words, x\(\ne-1\).
So what are you left with?
2x-3/x-3 :)
or \(\Large f(x)=\frac{(2x-3)}{(x-3)}, .... x\ne -1\)
What else do you see?
You can take out the x-3 and ur left with the 2
You cannot simplify further, so you have to think of graphing what's left.
Ok:)
What graphing features do you see in the simplified function?
ok hold on:)
It goes in opposite directions
http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJ4XjIiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjEwMDB9XQ--
Well, have you learned about asymptotes?
A horizontal asymptote for this may be 2/3
Horizontal asymptote is correct, but not at 3/2. Horizontal asymptote can be obtained by taking limits of f(x) as x->inf and x->-inf.
Do you see a vertical asymptote?
no! Okay thats wrong! I mean 3,-3
Almost! Vertical asymptotes are governed by the denominator becoming zero, so you need to find x for which (x-3)=0, or x=3. That's where the vertical asymptote is located.
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What about the horizontal asymptote?
hm...I give up on that one haha! So, I'm just going to guess and say 2/1
You would divide the coefficients of x, so 2x/x=2, yes 2/1=2 is correct.
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:) Yay! Now what about the domain and range?
We'll finish drawing the graph first. Do you know what it looks like?
A point of intersection? Between the vertical and horizontal lines
No this graph does not intersect itself, but it runs close to the two asymptotes that i drew.
To know which quadrant the graph goes, you would evaluate f(x) at a point just to the left of x=3 (say 2.9) and a point just to the right of x=3, say 3.1 and find out where to plot the lines. So can you find f(2.9) and f(3.1)?
=-0.2
not really, f(2.9)=(2*2.9-3)/(2.9-3)=2.8/(-0.1)=-28 Your turn to do the same for f(3.1)...
-32
It's +32. The denominator is +0.1.
So now we can complete the graph, remembering that \(x\ne-1\) !
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Can you now do the domain and range?
I'll try. So y>-1 and the domain could be all real numbers except 3
How about -1 which we have shown an empty circle?
Idk:(
We said to cancel the common factor (x+1), we need to put the restriction \(x+1\ne 0\) or \(x\ne -1\), so =1 cannot show up in the domain, right?
Right!
*-1 So what is the domain?
there would be one?
Your proposed domain was almost right, except that you missed out the -1, so...
Ok:) Thanks<3 I thought so, but I wasnt sure
So what is it?
all reals except 3,-1
Good! Now proceed with the range, and you'd be done!
Thank goodness lol!
So what's the range?
**Silence** Takes a wild guess and goes with...y>2/3
Look at the graph for a start! There are two points to be excluded.
2 and 3?
We're looking at range, or the y-axis. So look for y-values that we cannot get from the graph!
On the graph? ok so -1
-1 is for x. What is the corresponding value for y????
I.e. what is f(x) as x -> -1?
i gotta go eat dinner! Thanks so much for all your help:). I'll be back in 30 minutes or so:)
I will be gone by then. Someone else should be able to continue! Bon apetite!

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