Question

Write the rational function, f(x), which corresponds to the given characteristics of each graph.
5. zeros: x = 1
vertical asymptotes: x = -2
holes: none
horizontal asymptote: y = 1

6: zeros: none
vertical asymptotes: x = 3
holes: none
horizontal asymptote: y = 0

7. zeros: x = -1
vertical asymptote: none
holes: x = -4
horizontal asymptote: none

Answer 1

I'll show you how to do the first one and then you can try to do the others.
So I'm sure you know that a rational function is. To find the zeros of one, you have to think about where the top equation would be zero. In this case, it has a zero at x = 1. So the equation for this would be either \[1-x\] or \[x -1\] right? Or some variant because it would still work if it was multiplied by anything.

So you know the top of your equation is one of those two.

Then to find vertical asymptotes you have to find when the equation is being divided by 0, because then it will never reach that point, but it will get close as x gets infinitely small.
So \[2 + -2 = 0\] right?
So then, \[2+x = 0\] when x = -2.
See where I'm going?

Now you have \[\frac{1-x }{2+x } or \frac{x-1}{2+x}\]
If you plot both of those, you'll find that one of them fits all of the parameters you gave, whereas the other doesn't have a horizontal asymptote of 1.

If you have any questions, go ahead and ask.

Answer 2

what problem was that @A_Gnarly_Narwhal 5, 6 or 7

Answer 3

Sorry I should have explicitly mentioned it. It was the first one , or number 5.

Answer 4

a) zeros: none
b) vertical asymptotes: x = 3
c) holes: none
d) horizontal asymptote: y = 0

d) says the degree of the numerator must be less than the degree of the denominato
c) says there are no common factors top and bottom
b) says the denominator has a factor of \(x-3)\) and
a) says the the numerator has no zeros

simplest version could be \(\frac{1}{x-3}\)

Answer 5

@Mertsj Just #7

Answer 6

How could it have a zero of -1?

Answer 7

it means (-1, 0)

Answer 8

What I was hoping you had learned from those other 2 guys was that a rational function has a 0 if the numerator is 0.

Answer 9

So if x=-1 is a 0, then x+1 must be a factor of the numerator.

Answer 10

would the denominator be x + 4

Answer 11

If I remember right, a hole at -4 meant the function is undefined at -4 but it is a hole instead of an asymptote because the factor cancels out so I think it is:
\[\frac{(x+1)(x+4)}{x+4}\]

Answer 12

see I have a problem

Answer 13

Yep

Answer 14

the next problem has everything with it:
zero - x = -2
vertical asymptote - x = 1
hole - x = -3
horizontal asymptote - y = 1

Answer 15

So here is your chance to put it all together. Everything you have learned from the other three problems.

Answer 16

I got this so far: \[\frac{ (x + 2) (x + 3) }{ (x + 3) }\]

Answer 17

Is that right?

Answer 18

The vertical asymptote is x=1so there should be a factor of (x-1) in the denominator

Answer 19

I had it right, just I needed to add an (x - 1) in the denominator

Answer 20

Otherwise it looks good.

Answer 21

man I thought I would've gotten that one wrong

Answer 22

You surprised yourself...you are smarter than you thought.

Answer 23

@Mertsj My Answer: (x - 2) / (x - 2) (x - 4)

Problem:
Zeros: none
VA: none
Hole: x = 2
HA: y = 4

Answer 24

Is it right @Mertsj

Answer 25

Can't be because the function you wrote would have a zero of 2

Answer 26

Oh no. I just read about the hole.

Answer 27

so changed your answer

Answer 28

Also since the horizontal is y = 4, that means the degrees of numerator and denominator are the same but the ratio of the coefficients of the terms with the highest exponents is 4.

Answer 29

So make the numerator 4(x-2)

Answer 30

I have 1 more for you to check. @Mertsj My Answer: (x +/- 0) (x - 3/2) / (x - 1/2)

Problem:
Zero: x = 0
VA: x = 3/2
Hole: none
HA: y = 1/2

Answer 31

\[\frac{x}{(2x-3)}\]

Answer 32

so was my answer not simplified or something

Answer 33

The horizontal asymptote is x=1/2 which means the degrees of numerator and denominator are equal but the coefficients have ratio 1/2

Answer 34

So that horizontal asymptote does mean you have a factor in either numerator or denominator.

Answer 35

oh I see