Question

Identify the vertical asymptotes of f(x) = 10 over quantity x squared minus 7x minus 30

x = 10 and x = 3

x = 10 and x = -3

x = -10 and x = 3

x = -10 and x = -3

Help please im lost

Answer 1

\[\large \frac{10}{x^2 - 7x - 30}\]

Vertical Asymptotes occur when the denominator = 0

So when does \(\large x^2 - 7x - 30\) = 0?

Answer 2

would it be x=-10 and x=3 ? @johnweldon1993

Answer 3

Not quite! But close...

When you factor \(\large x^2 - 7x - 30\) is is true that you get

(x - 10)(x + 3)

But when do those = 0? when you plug in x = 10 and x = -3 there...you get 0

So it would be x = 10 and x = -3

Does that make sense?

Answer 4

yesss makes sense , help me on one more please (: @johnweldon1993

Answer 5

Sure :)

Answer 6

Identify the vertical asymptotes of f(x) = quantity x minus 4 over quantity x squared plus 13 x plus 36

x = -9 and x = -4

x = -9 and x = 4

x = 9 and x = -4

x = 9 and x = 4

@johnweldon1993

Answer 7

Alright so same thing here...

\[\large \frac{x - 4}{x^2 + 13x + 36}\]

So when does that denominator = 0?

\[\large x^2 + 13x + 36 = 0\]

Answer 8

would it be 9 & -4 ? @johnweldon1993

Answer 9

Hmm, when we factor that...we get what?

Answer 10

So lets see....

What 2 numbers when multiplied = 36 AND when added = 13?

Well 9 times 4 = 36 AND 9 + 4 = 13 so there we go

Factored, that would look like

(x + 9)(x + 4)

We just need to see when that = 0

Well if you have x = -9...you get 0....and when you have x = -4 you also get 0

so

x = -9 and x = -4

Answer 11

thank you so muchhhhhh . will you be on for long its just that im on this chapter that i honestly dont get and its online virtual school , so its not like i have a real teacher & your awesome at explaining this

Answer 12

Lol sure I'll stay on as long as I can for you :)

Answer 13

THANKSSSSS (: !
Identify the oblique asymptote of f(x) = quantity x plus 4 over quantity 3 x squared plus 5 x minus 2

y = 3x - 7

y = 3x + 17

y = 0

No oblique asymptote

-----------------------------------------

Identify the horizontal asymptote of f(x) = quantity 7 x plus 1 over quantity 2 x minus 9

y = 0

y = negative 1 over 9

y = 7 over 2

No horizontal asymptote

--------------------------------------------

Identify the oblique asymptote of f(x) = quantity 2 x squared minus 5 x plus 2 over quantity x minus 3

y = 0

y = 2x + 1

y = 2x - 11

No oblique asymptote

---------------------------------------------
Identify the horizontal asymptote of f(x) = 3 over 5 x

y = 3 over 5

y = 0

y = 5 over 3

No horizontal asymptote

ANY IDEA HOW TO SOLVE THIS :/

Answer 14

i just tried # 2 would that one be 7 / 2 ? if i did it right......

Answer 15

Alright

Oblique asymptotes are found using long division

Vertical asymptotes are found when the denominator = 0

Horizontal asymptotes are found when you see what the function equals when approaching infinity

Answer 16

So lets do number 1 first :)

Answer 17

i got No oblique asymptote

Answer 18

\[\large \frac{x + 4}{3x^2 + 5x - 2}\]

And you are correct saying there is NO oblique asymptote :)

When there is a higher exponent on the bottom than the top...there will not be an obique asymptote

Answer 19

Number 2

\[\large \frac{7x + 1}{2x - 9}\]
When you set x = infinity...you do INDEED get 7/2 as you showed :) good job!

Answer 20

What do you get for number 3?

Answer 21

thanks working #3 now would that one be y=2x+1

Answer 22

\[\large \frac{2x^2 - 5x + 2}{x - 3} \]

So lets see



2x + 1 would be correct! :)

Answer 23

yayyyyy im getting this & last one i got y=0

Answer 24

Andddddd that would make you perfect :)

Answer 25

thanks for the help !

Answer 26

No problem you did all that work lol :)