Question

f(x)=2/(x^2-2x-3) and find Domain and Range, x and y Intercept(s), Horizontal Asymptote(s), Vertical Asymptote(s)
i already graphed it but im having trouble finding the rest.

Answer 1

@Butterfly16 @satellite73

Answer 2

First, factor the denominator, because that's where all the action is:

x²-2x-3=(x+1)(x-3).

Now, when this is 0, you cannot calculate f(x), do the domain is all real numbers except for two. You know what I mean?

Answer 3

i think so

Answer 4

Denominator is 0 for x=-1 and x=3, so we don't want these in our domain

We now have:\[f(x)=\frac{ 2 }{ (x+1)(x-3) }\]and\[D _{f}=\mathbb{R} \backslash \{-1,3\}\]Do you recognize this notation for the domain of f?

Answer 5

yes

Answer 6

If you have already graphed it, you will have seen where the asymptotes are. Do you need to do calculations with limits, or can you just state the equations of the asymptotes?

Answer 7

state the equation

Answer 8

OK, can you see where the vertical asymptotes are?

Answer 9

https://www.google.com/search?q=State+the+horizontal+asymptote+of+the+rational+function.+f(x)+%3D(x%2B9)%2F(x%5E2%2B2x%2B3)&oq=State+the+horizontal+asymptote+of+the+rational+function.+f(x)+%3D(x%2B9)%2F(x%5E2%2B2x%2B3)&aqs=chrome.0.57j62l3.3298&sourceid=chrome&ie=UTF-8#hl=en&tbo=d&sclient=psy-ab&q=f(x)%3D2%2F(x%5E2-2x-3)&oq=f(x)%3D2%2F(x%5E2-2x-3)&gs_l=serp.3..0i8i30l3.975647.1007346.2.1007729.17.17.0.0.0.0.341.3078.0j12j1j4.17.0.les%3B..0.0...1c.1.dx78jvC6uk8&psj=1&bav=on.2,or.r_gc.r_pw.r_cp.r_qf.&bvm=bv.41248874,d.b2I&fp=e5351287a4c2a4a0&biw=1280&bih=685

Answer 10

Here's my graph. I also included a graph of another function: just y=(x+1)(x-3), which is the denominator of f. It's the blue graph. See the zeros of the blue one are the asymptotes of the red one?

Answer 11

BTW, you might want to give Geogebra a try. It's a great maths program (FREE!).
Look for it at geogebra.org.
I use it for all kinds of graphs.

Answer 12

No, that's not true, the asymptotes of the red graph are at the zeros of the blue, they are at x=-1 and x=3, just the numbers we couldn't use, because of the dividing by 0.

Answer 13

oh ok thanks

Answer 14

Try to see your function as 2 divided by the values of the parabola in the graph. That's why the zeros of the blue one are the asymptotes of the red (=your function).

Answer 15

ok

Answer 16

Now what have we left to do?

-intercept with x and y axes:

with x-axis: then f(x)=0. Now look at f:\[f(x)=\frac{ 2 }{ (x+1)(x-3) }\]
f is a fraction. A fraction is 0 if the numerator is 0. But: the numerator is (always) 2.
So the graph of f never intersects the x-axis. We already saw that...

Intercept with y-axis: just calculate f(0).

Answer 17

I've got f(0)=-2/3 so (0, -2/3) is intercept with y-axis.

What about the horizontal asymptote(s)?
Then we have to see what happens to f(x) for really very large (positive or negative) x.
Can you see what happens if you should make x very large (positive or negative)?

Answer 18

Come on, @mgarcia634, it's your turn ;)

Answer 19

positive then negative?

Answer 20

Just look at the graph to see what happens. It makes no difference if x is very large positive or very large negative. In other words: what happens to the graph if x is very far away from the origin?

Answer 21

Such values of x make the denominator a very large number. If you then divide 2 by that very large number (which is what f does), there is a very small number left. The graph tends to zero. Conclusion: the x-axis is the horizontal asymptote (equation: y=0).

Answer 22

i dont understand. how do i find the horizontal asymptote?

Answer 23

You find the horizontal asymptote by looking what happens with f(x) when you make x larger and larger.
f(10)=0.02597
f(20)=0.0056
f(50)=0.00083
smaller and smaller they get. f(x) tends to 0 as x goes to infinity.
The graph of f goes nearer and nearer to the x-axis. Therefore, the x-axis is the horizontal asymptote. Its'equation is y=0.

For negative values of x, the same behavior is observed, so that makes no difference.
Just take a look at the graph and you'll see what is going on.

Answer 24

is there no horizontal asymptotes? because If the degree of the denominator is less than the degree of the numerator then there is no Horizontal Asymptote.

Answer 25

Now there is only one thing left to do: what is the range of f?
Is it all real numbers? After all the graph goes all the way up and down along the asymptotes. But there is an area where the graph doesn't go: there is a maximum exactly between the asymptotes, so that is for x=1 (see graph). The value of this maximum is f(1)=-1/2.
Between y=-1/2 and the x-axis there is no part of the graph, so from all the real numbers, we must exclude that part:\[R _{f}=\mathbb{R} \backslash (-\frac{ 1 }{ 2 },0]\]

Answer 26

About your last question: the degree of the denominator is 2, that of the numerator is 0, so the degree of the denominator is greater than that of the numerator...

Answer 27

so domain is -1,3 and range is -1/2,0

Answer 28

No, domain is every real number, except two numbers:-1 and 3,
the range is every real number, except those from -1/2 to 0.
Mathematical notation:\[D_f=\mathbb{R} \backslash \{-1,3 \}\]\[R_f=\mathbb{R} \backslash (-\frac{ 1 }{ 2 }, 0]\]If you study the graph of f, you'll see both domain and range.