Double checking my work

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April 25th 2008, 03:05 PM
Sammyj

Double checking my work

I want to make sure I'm on the right path so far

R(x) = (x^2 -4)/(x^2)

Domain: x not equal to -2, x not equal to + 2
Range: all real
Vertical Asymptote: x = -2, x = 2
Horizontal Asymptote: y=1
Zeros: none

f(x) =x^3 - 2x^2 - 5x + 6

Domain: all real
Range: (5, + infinity)
Zeros: ?
April 25th 2008, 03:22 PM
elizsimca

Would you be so kind as to use parenthesis in your equations...I think you need to make some corrections perhaps, but I don't want to jump the gun. Parenthesis would make it more clear of what is on top of the fraction or on bottom.

Thanks!!

-Liz.(Hi)
April 25th 2008, 03:29 PM
Sammyj

Ok all done
April 25th 2008, 03:46 PM
elizsimca

Quote:

Originally Posted by Sammyj

I want to make sure I'm on the right path so far

R(x) = (x^2 -4)/(x^2)

Domain: x not equal to -2, x not equal to + 2
Range: all real
Vertical Asymptote: x = -2, x = 2
Horizontal Asymptote: y=1
Zeros: none

f(x) =x^3 - 2x^2 - 5x + 6

Domain: all real
Range: (5, + infinity)
Zeros: ?

$R(x)=\frac{x^2-4}{x^2}$

On this one, when x equals +/- 2, the result is zero. Therefore, it is defined at x=2 and x=-2. When x equals zero is where we have the problem. If you put zero into that equation you get $\frac{-4}{0}$ , hence your doman should be all real numbers not equal to 0. Because of this, your only vertical asymptote is at x=0.

Since the numerator and denominator are the same degree, to find the y-asymptote you divide the leading coefficients of the highest powers by each other, that is the equation for you line. In this case, the leading coeffiecients of the highest power, x^2, are 1 and 1. Dividing 1 by 1 yields 1. Therefore your horizontal asymptote is y=1. This should give you a hint about the range.

There are 2 zeros in this function. Set the top equal to zero and solve.

$x^2-4=0$
$x^2=4$
$x=\pm2$

On the second one I suggest using synthetic division to find the first zero. Your only possible zeroes from $\pm{\frac{p}{q}}$ are -1, 1, -6, 6, -3, 3, -2, 2. I think starting with 1 and -1 would be good. If you don't know what I'm talking about I can come back and explain! :)

After you use synthetic division and find the first zero, you will be left with a quadratic which you can solve by factoring or using the quadratic formula.

Good luck!
April 25th 2008, 03:54 PM
elizsimca

Also,

on the second problem. Your domain and range should be all real numbers. This is because x and y are defined for all values.

I've found that it really helps to graph these equations to get a feel for what you are looking at. Do you have a graphing calculator? It helps especially with polynomials with many possible zeroes so that you can try to pinpoint where to start with synthetic division.

:)
April 25th 2008, 04:06 PM
Sammyj

[quote=elizsimca;136009] $R(x)=\frac{x^2-4}{x^2}$

On this one, when x equals +/- 2, the result is zero. Therefore, it is defined at x=2 and x=-2. When x equals zero is where we have the problem. If you put zero into that equation you get $\frac{-4}{0}$ , hence your doman should be all real numbers not equal to 0. Because of this, your only vertical asymptote is at x=0.

Since the numerator and denominator are the same degree, to find the y-asymptote you divide the leading coefficients of the highest powers by each other, that is the equation for you line. In this case, the leading coeffiecients of the highest power, x^2, are 1 and 1. Dividing 1 by 1 yields 1. Therefore your horizontal asymptote is y=1. This should give you a hint about the range.

There are 2 zeros in this function. Set the top equal to zero and solve.

$x^2-4=0$
$x^2=4$
$x=\pm2$
[quote]

For the domain is it x no equal to +/- 2 or all real numbers?
For the range is it (1, infinity)

I hope I'm understand this right (Doh)
April 25th 2008, 04:12 PM
elizsimca

For the domain is it x no equal to +/- 2 or all real numbers?
For the range is it (1, infinity)

I hope I'm understand this right (Doh)[/quote]

The domain consists of all the possible x values. At x=+/-2, y=0. Therefore, the function has output values when you put in 2 or -2. This means that -2 and 2 are in the domain.

x=0 is the only value that does not have an output (or corresponding y-value.) Because of this, x=0 is the only number that should not be included in the domain. Therefore, the domain is all real numbers not equal to x.

You're close on the range. The range should be $(-\infty,1)$

This is because the function has y-values all the way down at negative infinity, and moving up to 1 (but not including 1 because y=1 is a horizontal asymptote).

Does this help?
April 25th 2008, 04:12 PM
Sammyj

Quote:

Originally Posted by elizsimca

Also,

on the second problem. Your domain and range should be all real numbers. This is because x and y are defined for all values.

I've found that it really helps to graph these equations to get a feel for what you are looking at. Do you have a graphing calculator? It helps especially with polynomials with many possible zeroes so that you can try to pinpoint where to start with synthetic division.

:)

Ok so if I did this right I got Zeros: x = -2, x = 1 and the range & domain are all real numbers (Worried)
April 25th 2008, 04:13 PM
elizsimca

And, I have to get ready to go to a play, but I will check back tonight to see how you are coming along!

Good luck!
April 25th 2008, 04:15 PM
Sammyj

Quote:

Originally Posted by elizsimca

For the domain is it x no equal to +/- 2 or all real numbers?
For the range is it (1, infinity)

I hope I'm understand this right (Doh)

The domain consists of all the possible x values. At x=+/-2, y=0. Therefore, the function has output values when you put in 2 or -2. This means that -2 and 2 are in the domain.

x=0 is the only value that does not have an output (or corresponding y-value.) Because of this, x=0 is the only number that should not be included in the domain. Therefore, the domain is all real numbers not equal to x.

You're close on the range. The range should be $(-\infty,1)$

This is because the function has y-values all the way down at negative infinity, and moving up to 1 (but not including 1 because y=1 is a horizontal asymptote).

Does this help?

ahhh this makes some sense now (Clapping)
April 25th 2008, 07:44 PM
elizsimca

Quote:

Originally Posted by Sammyj

Ok so if I did this right I got Zeros: x = -2, x = 1 and the range & domain are all real numbers (Worried)

I don't know how to synthetically divide using the latex code..you have 2 correct zeroes, but you are missing one...I will try to show you:

If you synthetically divide 1 into the coefficients of the polynomial, you'll end up with a remainder of:

1 -1 -6

This means that you can say that $x^3-2x^2-5x+6$ is equal to $(x-1)(x^2-x-6)$

Then you can factor the remaining quadratic:

$x^2-x-6=(x-3)(x+2)$

So altogether, you have $x^3-2x^2-5x+6=(x-1)(x-3)(x+2)$ .

Setting each of the factors equal to zero yields:

x=1
x=3
x=-2

Hopefully this helps!