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

Answer 1

\[f(x) = \frac{ 10 }{ x^2 -7x -30 }\]

Answer 2

set \[x^2-7x-30=0\] solve for \(x\)

Answer 3

okay give me a sec

Answer 4

fortunalely for you this factors as
\[(x-10)(x+3)=0\] so the zeros are easy to find

Answer 5

yeah @satellite73 gave you a huge hint there

Answer 6

so its x = -10 and x = 3

Answer 7

no dear

Answer 8

\[x-10=0\\
x=10\]

Answer 9

also
\[x+3=0\\
x=-3\]

Answer 10

oh so i have to bring it to the other side okay i see

Answer 11

riiiiiigght

Answer 12

thnx i have one more do you mind helping with?

Answer 13

sure

Answer 14

Identify the horizontal asymptote of f(x) = quantity 2 x minus 1 over quantity x squared minus 7 x plus 3.

y = 0
y = 1 over 2
y = 2
No horizontal asymptote

Answer 15

\[f(x) = \frac{ 2x - 1 }{ x^2 - 7x +3 }\]

Answer 16

do i basically do the same thing as before?

Answer 17

no!!

Answer 18

horizontal not vertical

Answer 19

oh so what do i need to do?

Answer 20

the degree of the numerator is less than the degree of the denominator

Answer 21

the degree of the numerator is 1, the degree of the denominator i s2

Answer 22

that means the denominator grows faster than the numerator so the horizontal asymptote is
\[y=0\]

Answer 23

thank you so much!!!

Answer 24

i gtg but i have lie 5 more questions to do later can you help with?

Answer 25

sure dear if i am on