Question

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

Answer 1

@campbell_st

Answer 2

10 / (x^2-7x-30)

wont the asymptote be formed by finding out what minimum value x can be and still be true in the equation: x^2 - 7x - 30, because we can't take the square of a negative number, right?

Answer 3

\[10 \over x^2 - 7x - 30\]

Answer 4

Yeah we can't square a negative

Answer 5

also, these are my answer choices if it helps...
x = 10 and x = 3

x = 10 and x = -3

x = -10 and x = 3

x = -10 and x = -3

Answer 6

so take the equation part x^2 -7x -30 set it equal to zero, and solve for the roots

make sense?

Answer 7

yeah

Answer 8

so what two number can we add to make -7 or
p + q = -7

and what two numbers timesed by each other makes -30 or
p*q = -30

Answer 9

15(-2)

and... -2(3.5)

Answer 10

maybe if I just did it this way it'd be easier for you

(x-10)(x+ ? ) = x^2 -7x -30

Answer 11

x + 10?

Answer 12

if we foiled:
(x-10)(x+10)
we'd get:
x^2 -100

that dones't equal x^2 - 7x -30 that we need.

try one more time, then i'll step in ^_^

Answer 13

hmm... 3 lol?

Answer 14

haha ya!

(x-10)(x+3)

so then we find the roots by this method thing... here:
(x-10) = 0, solve for x
(x+3) = 0, solve for x

Answer 15

x = 10 and x = -3

Answer 16

yep!
those are the vertical asytompes

Answer 17

Thanks!