Question

12.) The Rational Function Has a y-Intercept of 7. What Is The Equation For This Function?

Please Help, I don't Understand?!

Answer 1

Please Help!! @phi ?

Answer 2

Someone please??? Does anyone understand this??? @whpalmer4 @phi ???

Answer 3

Okay, do you have any idea what the general form of the equation is?

Answer 4

can you identify the asymptotes from the diagram?

Answer 5

No, I Have No clue :(

Answer 6

Can you give me the equations for the dashed lines on the diagram? Those the asymptotes

Answer 7

x=-2, y=-5?

Answer 8

oops y=5

Answer 9

@whpalmer4 ? you there?

Answer 10

:(

Answer 11

yes, those are correct. Do you understand what features of an equation would cause such asymptotes?

Answer 12

No I dont..

Answer 13

Okay, I was afraid of that, but I guess I'll just have to teach you the material :-)

Let's take a simple case. \[y = \frac{1}{x}\]Graph looks like this:

Answer 14

Is there a value of \(y\) at \(x = 0\)? If so, what is it? If not, why not?

Answer 15

Wait, why would the graph look like that? What makes the points be that way?

Answer 16

What is 1/0 ?

Answer 17

It's zero

Answer 18

Ah, so 0*0 = 1?

Answer 19

What is 2/0?

Answer 20

No, hmm, Im confused ahha

Answer 21

if \(a/b = c\) then \(c*b = a\)

Answer 22

that implies that division by 0 is a no-no, because the only number we could divide by 0 would be 0 to have that be true.

Answer 23

10/5 =2 because 2*5 = 10

So, we can't meaningfully divide by 0. There's a spot right there in the middle of the graph where we don't have a value!

Now let's investigate a bit more. What if we have x = 2? y = 1/x = ?

Answer 24

@MissLovelyPearlOyster are you going to respond?

Answer 25

sorry my teacher called me, just a minute lemme read all this..

Answer 26

y=1/2? Is that were the curve starts in the graph?

Answer 27

@whpalmer4 ?

Answer 28

yes. if x=2, y = 1/2. how about if x = 1/2?

Answer 29

Idk, other than using a decimal on the bottom hah y=1/0.5?

Answer 30

well, you should know how to divide by a fraction by now! Invert the fraction from the denominator and multiply by it instead:

\[\frac{1}{\frac{1}{2}} = 1* \frac{2}{1} = 2\]

Answer 31

how about if x = 1/100, what is y?

Answer 32

y=100?

Answer 33

yep. so, as we get closer and closer to x = 0, y gets bigger and bigger, right?

Answer 34

yes it does..

Answer 35

Okay. what if we have x = -1? y =?

Answer 36

y=-1?

Answer 37

Yep. How about x = -1/2? x = -1/100?

what happens as x gets closer and closer to x = 0, but from the negative side?

Answer 38

y=-2, y=-100 ? and y gets smaller?

Answer 39

it goes bigger and bigger, but in the negative direction. |y| increases, right?

Answer 40

yes y gets smaller

Answer 41

i mean bigger hahah

Answer 42

the magnitude of y gets larger...y becomes more and more negative. so on the left side of x = 0, y -> -infinity as we approach x = 0, and on the right side of x = 0, y -> + infinity as we approach x = 0

Answer 43

okay..

Answer 44

and that's exactly what the graph shows, of course. at x=0, y has to be in an infinite number of places at once :-) now, because y approaches that line x = 0 ever and ever closer without actually getting there, we call that an asymptote
a vertical asymptote in this case

Answer 45

a rational function will have a vertical asymptote wherever the denominator = 0.

Answer 46

our rational function is \[y = \frac{1}{x}\]there is a vertical asymptote at \(x = 0\) because the denominator is 0 there.

Clear?

Answer 47

What if we wanted a similar graph, except with a vertical asymptote at \(x = 1\)?

Answer 48

hmm.. I honestly don't understand all you're big words ahaha

Answer 49

do you understand why we have a vertical asymptote at x = 0?

Answer 50

because y doesnt get to that point where x is 0?

Answer 51

as we get closer and closer to x = 0, y gets closer and closer to the line x = 0, and if you tell me just how close you want it, I can tell you a spot where that is true. and it has a discontinuity where x = 0, because at x = 0-a little bit, y is negative, and at x = 0+a little bit, y is positive, and at no point does y ever equal 0. It has to "snap" from positive to negative, or the other way around. We call that a discontinuity, where there isn't a smooth connection.

Answer 52

we get one at any value of \(x\) that makes our denominator = 0.

Answer 53

how could we write a function that has a denominator that = 0 when x = 1?

Answer 54

How about \[y = \frac{1}{x-1}\]?

Answer 55

what is the value of \(x\) that makes the denominator = 0?

Answer 56

Wait a minute, lemme watch a video on this okay? When I come back, lemme see if i can answer you're questions :)

Answer 57

okay. btw, "you're" = "you are", "your" is the possessive, as in "your notebook"

Answer 58

I know, auto correct, I totally messed up that word haha I'm only stupid in math, but a genius in English. Omg this is still confusing me, and I'm not really comprehending what you're saying. Just try and explain the problem

Answer 59

start with reading the book :D

Answer 60

well, let's let it go at this: if we want a vertical asymptote at some value of x, we need to arrange for the denominator to equal 0 when x equals that value where we want the vertical asymptote.

Answer 61

Hahah my text book doesn't really cover this, they cover how to get the graph, not get the equation from the graph

Answer 62

if we want the denominator to equal 0 at x = -2, how about we use
\[y = \frac{1}{x+2}\]?

Answer 63

Looks like this:

Answer 64

(ignore the vertical line at x = -2, that's an artifact of the graphing software)

Answer 65

Looks like our graph, except not quite high enough up on the page, right?

Answer 66

yes it does.. where does the x=-2 and y=5 go in the equation? Those are the asymptotes right?

Answer 67

well, they don't really go in the equation — the asymptotes are values that are never touched (in the case of vertical asymptotes), only approached

Answer 68

if we put the \(x=-2\) into our formula:
\[y = \frac{1}{x+2} = \frac{1}{-2+2} = \frac{1}{0}\]

Answer 69

Ohh okay, see now this makes a tiny bit of sense

Answer 70

So what does that have to do with the equation?

Answer 71

that's our vertical asymptote.

Answer 72

ohhh because the -2 is the x asymptote, so we would plug that in for x? And it must equal to zero, so the denominator must be x+2 so the -2 crosses it out. Am I right so far?

Answer 73

yes. to put a vertical asymptote at x = a, we need a term in the denominator: \((x-a)\)

If we need multiple vertical asymptotes, we have a product in the denominator: \[(x-a)(x-b)(x-c)\]

Answer 74

Remember that \(a,b,c\) may be positive or negative, so there may be + signs in the final equation, not just -.

Answer 75

but we dont need multiple vertical asymtotes right

Answer 76

is the denominator finished? y=1/(x+2) ? what about the top now?

Answer 77

sorry if im rushing you, I have to leave soon and really need this turned in

Answer 78

denominator is finished.

Answer 79

The top/ how would I do that? You're a great teacher btw

Answer 80

I'm still trying to figure out how to get the top and y-intercept...

Answer 81

Okay, I think I have it.

We can add 5 to the result of our fraction to shift the graph up by 5, right?

\[y = \frac{1}{x+2}+5\]looks like

Answer 82

Now, we need to adjust the sharpness of the "corner" so that the curve goes through the point \((0,7)\) We can do that by putting a value other than 1 in our numerator:

Answer 83

If we make our function \[y = \frac{a}{x+2}+5\]and find the value of \(a\) so that the curve goes through \((0,7)\) we'll be all set.

\[7 = \frac{a}{0+2}+5\]\[7 = \frac{a}{2}+5\]\[a=\]

Answer 84

so the numerator is 12?

Answer 85

No. try solving that again.

Answer 86

ohh a=4

Answer 87

Yes. So I believe our function is\[y = \frac{4}{x+2}+5\]
It goes through our y-intercept of (0,7):
\[y = \frac{4}{0+2}+5 = 2+5 = 7\]
and it has the right asymptotes.

Answer 88

We took the graph of \(y = f(x) = \frac{1}{x}\) and converted it to \[y=4f(x+2)+5\]which gives it a vertical stretching by a factor of 4, shifts it up 5, and left 2.

Answer 89

OMG thank you!! I understand this now :) God bless you, and now I have to log off. Have a great day!