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

- anonymous

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

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- anonymous

@mathmate

- anonymous

no hold on :)

- anonymous

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## More answers

- mathmate

Do you mean \(\Large f(x)=\frac{2x^2−x−3}{x^2−2x−3}\)

- anonymous

yes:)

- mathmate

Do you know how to factorize quadratic expressions?

- anonymous

yes!

- anonymous

So I got for the first one, (x+1)(2x-3)

- mathmate

So factor the numerator and denominator, as a start! ok?

- anonymous

Got it:) So, I gottt
(x+1)(2x-3)/(x+1)(x-3)

- mathmate

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?

- anonymous

Canceling out the x+1

- mathmate

Just like that, or with a condition?

- anonymous

I dont know what you mean by that? So, I'm going w/ "just like that."

- mathmate

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\(\ne-1\).

- mathmate

So what are you left with?

- anonymous

2x-3/x-3 :)

- mathmate

or
\(\Large f(x)=\frac{(2x-3)}{(x-3)}, .... x\ne -1\)

- mathmate

What else do you see?

- anonymous

You can take out the x-3 and ur left with the 2

- mathmate

You cannot simplify further, so you have to think of graphing what's left.

- anonymous

Ok:)

- mathmate

What graphing features do you see in the simplified function?

- anonymous

ok hold on:)

- anonymous

It goes in opposite directions

- anonymous

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJ4XjIiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjEwMDB9XQ--

- mathmate

Well, have you learned about asymptotes?

- anonymous

A horizontal asymptote for this may be 2/3

- mathmate

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.

- mathmate

Do you see a vertical asymptote?

- anonymous

no! Okay thats wrong! I mean 3,-3

- mathmate

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.

- mathmate

|dw:1439339170261:dw|

- mathmate

What about the horizontal asymptote?

- anonymous

hm...I give up on that one haha! So, I'm just going to guess and say 2/1

- mathmate

You would divide the coefficients of x, so 2x/x=2, yes 2/1=2 is correct.

- mathmate

|dw:1439339327350:dw|

- anonymous

:) Yay! Now what about the domain and range?

- mathmate

We'll finish drawing the graph first.
Do you know what it looks like?

- anonymous

A point of intersection? Between the vertical and horizontal lines

- mathmate

No this graph does not intersect itself, but it runs close to the two asymptotes that i drew.

- mathmate

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)?

- anonymous

=-0.2

- mathmate

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)...

- anonymous

-32

- mathmate

It's +32. The denominator is +0.1.

- mathmate

So now we can complete the graph, remembering that \(x\ne-1\) !

- mathmate

|dw:1439339824439:dw|

- mathmate

Can you now do the domain and range?

- anonymous

I'll try. So y>-1 and the domain could be all real numbers except 3

- mathmate

How about -1 which we have shown an empty circle?

- anonymous

Idk:(

- mathmate

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?

- anonymous

Right!

- mathmate

*-1
So what is the domain?

- anonymous

there would be one?

- mathmate

Your proposed domain was almost right, except that you missed out the -1, so...

- anonymous

Ok:) Thanks<3 I thought so, but I wasnt sure

- mathmate

So what is it?

- anonymous

all reals except 3,-1

- mathmate

Good!
Now proceed with the range, and you'd be done!

- anonymous

Thank goodness lol!

- mathmate

So what's the range?

- anonymous

**Silence**
Takes a wild guess and goes with...y>2/3

- mathmate

Look at the graph for a start!
There are two points to be excluded.

- anonymous

2 and 3?

- mathmate

We're looking at range, or the y-axis. So look for y-values that we cannot get from the graph!

- anonymous

On the graph? ok so -1

- mathmate

-1 is for x. What is the corresponding value for y????

- mathmate

I.e. what is f(x) as x -> -1?

- anonymous

i gotta go eat dinner! Thanks so much for all your help:). I'll be back in 30 minutes or so:)

- mathmate

I will be gone by then. Someone else should be able to continue!
Bon apetite!

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