Question

Pls help

Answer 1

factoring the denominator and setting it to zero is a good start:

So:
\(4 x^2 + 18 x = 0\)

Answer 2

2x(x+9)

Answer 3

\(2x(x+9) = 2x^2 + 18x\)

Answer 4

ye

Answer 5

\[f(x)=\frac{ 2x^2+11x+9 }{ 4x^2+18x }\]
\[=\frac{ 2x^2+2x+9x+9 }{ 2x(2x+9) }\]
\[=\frac{ 2x(x+1)+9(x+1) }{ 2x(2x+9) }\]
\[=\frac{ (x+1)(2x+9) }{ 2x(2x+9) }\]
\[=\frac{ x+1 }{ 2x }\]
vertical assymptote is 2x=0
or
x=0

Answer 6

factor the numerator and denominator
cancel the common factor if any
set the denominator =0
and find the value of x