Question

For the three function I, II, and III attached inside, which match the following definitions? There can be one that matches, two, all three, or none at all.

a) Horizontal asymptote of y=4
**my answer II and III. is that right? :)

b) The x-axis is a horizontal asymptote
**not really sure which ones match this definition :/

c) Symmetric about the y-axis
**also don't get this one :(

d) An odd function
**how would i figure out which ones are odd or not odd?

e) Vertical asymptotes at x= +/- 4
**not sure which functions match this definition either :( but you factor and set to zero and solve for x right? :/ Thanks!! :)

Answer 1

three functions! :)

Answer 2

To get horizontal asymptotes, try to see how the functions behave as they approach plus or minus infinity...

Answer 3

a) Horizontal asymptote of y=4
**my answer II and III. is that right? :)

looks right

Answer 4

oh i thought it if the numerator has the highest exponent, theres no horizontal asymp. and if the denominator has it, then y=0? and if the exponents are equal, you divide the coefficients? :/ are those the correct rules?

Answer 5

oops... I really need to read the entirety of the stuff... yeah, tis right

Answer 6

ohh okay hahaha awesome :) yay! not really sure what part b is asking though :/

Answer 7

oh, and what it means for the x-axis to be a horizontal asymptote is if the line y = 0 is an asymptote.

Answer 8

ohh okay... how do you determine that?

Answer 9

b) The x-axis is a horizontal asymptote
means y=0 as terenz said. I think if you can find the others to be y=4, you can find this.

Answer 10

ohh I'm just finding the asymp. here?

ooh okayy... umm II and III, y=4
I denominator has the highest exponent, so y=0 ?

so for this one, function I is a horizontal asymptote?

Answer 11

Yes indeed, but only because the highest exponent is in the denominator, which means it becomes bigger much more quickly than the numerator, thus making the entire fraction arbitrarily small as x runs to infinity.

Answer 12

oooh okay yay! :) so onto c now! again, not sure what they mean by symmetric about the y-axis :/

Answer 13

For a function to be symmetric about the y axis, this MUST hold:

f(x) = f(-x)

Answer 14

okay.. ermm how do we figure that out? :/ not sure where to start

Answer 15

Well, try the first one, and replace all the x's with -x, and see if you get EXACTLY the same function.

Answer 16

ohh okie :)

I. y=4(-x)-1 -4x-1
_________ ____________
(-x)^2 + 16 = x^2 + 16

so this is a no right?

Answer 17

did i do that right? haha

Answer 18

Yes you did, is it EXACTLY the same function?

Answer 19

II. y=4(-x)^2-1 4x^2-1
_______________ = _____________
(-x)^2+16 x^2 + 16

ermm not sure if i did this right haha.. but if i did, this would be a yes right? since it's the same?

Answer 20

II is a yes, yes, (LOL)
What about III?

Answer 21

ooh yay! no it's not.. so function I. doesn't match the definition of c ?

Answer 22

hahaha

umm III. y=4(-x)^2+1 4x^2+1
_________________ = ____________
(-x)^2 - 16 x^2 - 16

so it's the same too right?

so II and III are the ones that match the definition? :)

Answer 23

That's right :>

Answer 24

ooh yay!! :)

Answer 25

Now the next one (odd function) is a bit tricky...
For a function to be odd, this MUST hold:

f(-x) = -f(x)

IE, when you replace all the x's with -x, the result is the NEGATIVE of the original function.

A rather simple odd function is x^3, or perhaps sin(x)

since (-x)^3 = -x^3 and sin(-x) = -sin(x)

Answer 26

By the way, another term for a function symmetrical about the y-axis is EVEN function ^_^

Answer 27

ohh okay... let's see if i'm able to do this right haha

I. f(-x)=4(-x)-1/(-x)^2+16 = -4x-1 / x^2 + 16

would this be odd? since it became -4x ?

Answer 28

Remember that the ENTIRE function must have been turned negative... that -1 in the numerator didn't become +1 now, did it? ;)

Answer 29

II. f(-x)=4(-x)^2 - 1 / (-x)^2 + 16 = 4x^2-1/ x^2 + 16

so this is a no right?

and ohh okay... so function I is also a no go for this definition?

Answer 30

no go, yup.

Answer 31

III. f(-x)=4(-x)^2 + 1 / (-x)^2 + 16 = 4x^2 + 1 / x^2 + 16

so this is a no too?

so this one would be none match right?

Answer 32

yup.

So...answer?

Answer 33

none ?

Answer 34

That's right ^_^

Next time this comes up, do yourself a favour, save some time:
An even function is automatically NOT an odd function (obviously)
and we just showed that II and III show even functions.

Answer 35

ohhh okie :) cool will keep that in mind! hehe so last part! :O

do you factor the bottom? :/

Answer 36

That said, just because a function is not even doesn't mean it's odd (compare with real numbers, just because a number is not even doesn't mean it's odd, take 0.5 for instance ;) )

Now, back to reality:
Just see if the function goes to plus or minus infinity as x goes to +/- 4.

Hop to it, now ^_^

Answer 37

haha okie :P

so i'm plugging in -4 and +4 into the x values?

Answer 38

I suppose you could do that... does it become infinity?

Answer 39

if so,

I. 4(4)-1 / (4)^2 +16= 15/32 ?
4(-4) - 1 / (-4)^2 + 16 = -16-1 / 16+16 = -17/32

?

so no, not infinity? or do you mean infinity as in an irrational number?

Answer 40

infinity def does not mean an irrational number :P

Means something divided by zero

Answer 41

Yes, that ^
Just make sure that it isn't zero over zero, as that warrants further investigation ^_^

Answer 42

okay.. so function I is not over 0, so this isn't an answer right?

Answer 43

Yup ^_^

Answer 44

II. 4(4)^2-1 / (4)^2+16 = 65/32
4(-4)^2-1 / (-4)^2 + 16= 65/32

so also not answer?

Answer 45

Yup.

Answer 46

III. 4(4)^2 + 1/(4)^2 -16 = 65/0
4(-4)^2 + 1 / (-4)^2 -16=65/0

? so the only one that matches definition e is function III ?

Answer 47

Ergo...?
;)

Answer 48

ergo... function III is the answer for letter e ? :P

Answer 49

Et voila ^_^

Good job there :)

Answer 50

yay!! thanks a bunch!!! :)

Answer 51

Pas de problème :)

Answer 52

haha are you taking french? (i think lol)

Answer 53

No :3

Answer 54

oh haha well your french is tres bon :P

Answer 55

Merci :)