Question

The figure below shows the graph of a rational function f with vertical asymptotes x=-2,x=-6,and horizontal asymptote y=-3. The graph also has x-intercepts of -4 and 1, and it passes through the point 0,1.

The equation for f(x) has one of the five forms shown below. Choose the appropriate form for f(x), and then write the equation. You can assume that f(x) is in simplest form.

A medal will be given for solving.

Answer 1

@Auxuris can you show me how to do this?

Answer 2

@ganeshie8 can you show me how to do this?

Answer 3

Since u have TWO "vertical asymptotes",
there are TWO values that can make the function go infinity

that means, there will be atleast TWO factors in the denominator.

Answer 4

Having that knowledge, u can discard two options straight away. wat wud be they ?

Answer 5

a&b

Answer 6

Yes !

Answer 7

next, use below to figure out the answer :-

since there are TWO x-intercepts,
when u set the rational function to ZERO,
there must be atleast TWO factors left in the NUMERATOR.

Answer 8

so that gets rid of c as well

Answer 9

that gets rid of c and d also

Answer 10

\(a\) is just a constant... so its not counted as factor for x-intercept okay ?

Answer 11

two x-intercepts means,
u must have atleast TWO factors of form : \((x-b)(x-c)\) in the numerator

Answer 12

so, answer is the last option

Answer 13

the answer is e. thank you

Answer 14

np.. u wlc :)

Answer 15

oh you already told me... lol thank you

Answer 16

good , u saw it first, before seeing my reply :)

Answer 17

good job !!

Answer 18

now, how do I figure out the equation?

Answer 19

just plugin the numbers.

Answer 20

\(\large f(x) = \frac{a(x-b)(x-c)}{(x-d)(x-e)}\)

Answer 21

\(d\) and \(e\) are the values which give u Vertical asymptotes right ?

Answer 22

The figure below shows the graph of a rational function f with vertical asymptotes x=-2,x=-6

Answer 23

so,
\(d = -2\)
\(e = -6\)

Answer 24

plugin them, the equation becomes :

\(\large f(x) = \frac{a(x-b)(x-c)}{(x--2)(x--6)}\)

\(\large f(x) = \frac{a(x-b)(x-c)}{(x+2)(x+6)}\)

Answer 25

fine, so far ?

Answer 26

I was with you until we started plugging the numbers in
Im getting but slowly, if that makes sense

Answer 27

ahh okay...
look at the above equation after plugging in numbers, see that \(x = -2\) and \(x = -6\) will make the denominator \(0\).

Answer 28

if that makes any sense.. .

Answer 29

you're doing a really good job explaining it im just not getting it

Answer 30

its okay... takes time....

you okay wid below statement :
"vertical asymptote occurs when the denominator becomes 0"

Answer 31

?

Answer 32

for example :
\(\large f(x) = \frac{1}{x-1}\)

will have an vertical asymptote at \(x=1\), cuz \(x = 1\) will make the denominator \(0\)

Answer 33

another example :
\(\large f(x) = \frac{1}{x}\)

will have a vertical asymptote at \(x=0\) cuz \(x=0\) will make the denominator \(0\)

Answer 34

u wana see why \(x=0\) will give a vertical asymptote for above example function ?

Answer 35

im reading it and I'm also reading other examples that are saying the same thing you are....its me, its not you, lol

Answer 36

good, im happy if u get the concept smoothly... :)

Answer 37

are you telling me to make the denominator 0?

Answer 38

hmm

Answer 39

this video explains nicely about vertical asymptotes :
https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/asymptotes-graphing-rational/v/asymptotes-of-rational-functions

Answer 40

ugh its not allowing me to pull up the video

Answer 41

oh no, wait

Answer 42

its opening fine for me... maybe some browser stuff... leave it -.-

Answer 43

I think I have it

Answer 44

good ^_^

Answer 45

(x+2)(x+6)=0?

Answer 46

yes, setting the bottom to 0, gives u vertical asymptotes.

Answer 47

how do I find the top values?

Answer 48

top values are x-intercepts, u wil get thm by setting top to 0

Answer 49

The graph also has x-intercepts of -4 and 1

Answer 50

so, setting top equal to 0, must give u :
x = -4 and x = 1

Answer 51

\[\frac{ \left( x+4 \right)\left( x-1 \right) }{\left( x+2 \right) \left( x+6 \right) }\]

Answer 52

yes, and u forgot a constant... \(a\) in the numerator.
u must keep it to account for the y-intercept.

Answer 53

\(\large f(x) = \frac{ a\left( x+4 \right)\left( x-1 \right) }{\left( x+2 \right) \left( x+6 \right) }\)

Answer 54

u still need to find the value of \(a\)

Answer 55

LOL! I did, but I didn't. I was trying to figure out how to find the constant

Answer 56

il give a hint : \((0, 1)\) is a point on the graph

Answer 57

so, the function must satisfy x = 0, y = 1

Answer 58

\(\large y = f(x) = \frac{ a\left( x+4 \right)\left( x-1 \right) }{\left( x+2 \right) \left( x+6 \right) } \)

put x = 0, y = 1

Answer 59

\(\large 1 = \frac{ a\left( 0+4 \right)\left( 0-1 \right) }{\left( 0+2 \right) \left( 0+6 \right) } \)

Answer 60

solve \(a\)

Answer 61

-3

Answer 62

\(\large \color{red}{\checkmark }\)

Answer 63

-3\[\frac{-3 \left( x+4 \right)\left( x-1 \right) }{ \left( x+2 \right)\left( x+6 \right) }\]