Question

help find the equation

Answer 1

what is the equation of the graph???

Answer 2

@Hero @Abhisar @ganeshie8

Answer 3

so the graph is behaving crazily at x = 3

Answer 4

see this graph
https://www.desmos.com/calculator/xulxmyhnxm

Answer 5

yea

Answer 6

how did you get the equation though

Answer 7

@ganeshie8

Answer 8

It was an educated guess. How is an asymptote related to the denominator of a rational function ?

Answer 9

well there is a horiz retricem at y=0 and there is a vert retricem at x=3

Answer 10

from your graph it lloks like y goes to \(-\infty\) at x=3.

can you think of a way to achieve this?

Answer 11

*looks

Answer 12

i.e. how can you get a function that goes to \(-\infty\) when \(x=3\)?

Answer 13

if the function is negative.?

Answer 14

y=-10

is a negative function but it does not go to \(-\infty\) at x=3

Answer 15

think of it this way - what would f(x) have to be for y to be equal to \(\infty\) in this function:\[y=\frac{1}{f(x)}\]

Answer 16

I meant what would the value of f(x) need to be in the above?

Answer 17

positive

Answer 18

e.g. if f(x)=1 then:\[y=\frac{1}{f(x)}=\frac{1}{1}=1\]think of what value f(x) should be in order to get y to be infinite

Answer 19

0

Answer 20

good :)

Answer 21

now think of a function that will be equal to zero when x=3, can you think of such a function?

Answer 22

e.g. if:\[f(x)=x-1\]then this will be equal to zero when x=1

Answer 23

if the bottom is 3

Answer 24

so x-3

Answer 25

perfect! :)

Answer 26

so now we know that if we have:\[y=\frac{1}{x-3}\]then this will be equal to infinity at x=3.

however, in your graph you want y to be negative infinity at x=3.

so what do you think we should do?

Answer 27

put a negative sign in front

Answer 28

great!

so we now have:\[y=-\frac{1}{x-3}\]we are't quite there yet...

Answer 29

the next thing to notice is that in your graph the value of y is ALWAYS negative - agreed?

Answer 30

yea

Answer 31

however the function we came up with:\[y=-\frac{1}{x-3}\]is positive if x<3 and negative if x>=3

Answer 32

so we need to think of a way to ensure that the value of the denominator (i.e. \(x-3\)) is ALWAYS positive no matter what the value of x is

Answer 33

okay

Answer 34

one way of doing this is to square it - so we get:\[y=-\frac{1}{(x-3)^2}\]

Answer 35

another way is to take its absolute value, i.e.:\[y=-\frac{1}{|x-3|}\]

Answer 36

yet another way would be to raise it to the 4th power, i.e.:\[y=-\frac{1}{(x-3)^4}\]etc

Answer 37

does that make sense now?

Answer 38

oh okay i understand

Answer 39

great! :)