Find the vertical asymptote(s) of f(x)= x^2 +3x+6/ x^2 -4

Hello,

f(x) = 2/x² + 3x - 10

Vertical asymptote is x=0

f(x) = x² + 6x -9/x - 2

No oblique asymptote.

= = = = = = = = = = = = = = = = = =

Indeed the problem would have been more interesting if you mattered enough to add some parenthesis...

= = = = = = = = = = = = = = = = = =

EDIT: ...so next time, bother a little to add the brackets where they need to be:

f(x) = 2/(x² + 3x - 10) = 2 / [(x + 5)(x - 2)]

Vertical asymptotes are x=2 and x=-5.

f(x) =(x² + 6x - 9) / (x - 2)

  = (x² - 2x + 8x - 16 + 7) / (x - 2)

 = x + 8 + 7/(x - 2)

Oblique asymptote is y=x+8

Precisely,

Step-by-step explanation:

0

Step-by-step explanation:

1.4

2.3

step-by-step explanation:

the formula for the constant of variation is

y= kx

divide each side by x

y/x =kx/x

y/x =k

we know a and y

1.   y/x = k

4/1 =4

check the other values to make sure they are the same

8/2 = 4   12/3 =4   and 16/4 =4  

the constant of variation is 4

2.   y/x = k

3/1 =3

check the other values to make sure they are the same

6/2 = 3   9/3 =3   and 12/4 =3  

the constant of variation is 3

c

step-by-step explanation:

at least i think its