Pre Calc question please help asap?

[Jordan]

Find the horizontal asymptote of f (x) = 6 (x+4) (3x-1) / (8-x) ( 2x+2)

y=__________

please explain

[youngandinsane]

horizontal asymptote is anywhere where the slope=0
slope=rise over run so rise (or the y value) has to be equal to 0

0=(6x+24)(3x-1)/(8-x)(2x+2)
and you just do all the algebra from there

[youngandinsane]

horizontal asymptote is anywhere where the slope=0
slope=rise over run so rise (or the y value) has to be equal to 0

0=(6x+24)(3x-1)/(8-x)(2x+2)
and you just do all the algebra from there

[youngandinsane]

horizontal asymptote is anywhere where the slope=0
slope=rise over run so rise (or the y value) has to be equal to 0

0=(6x+24)(3x-1)/(8-x)(2x+2)
and you just do all the algebra from there

[youngandinsane]

horizontal asymptote is anywhere where the slope=0
slope=rise over run so rise (or the y value) has to be equal to 0

0=(6x+24)(3x-1)/(8-x)(2x+2)
and you just do all the algebra from there

[youngandinsane]

horizontal asymptote is anywhere where the slope=0
slope=rise over run so rise (or the y value) has to be equal to 0

0=(6x+24)(3x-1)/(8-x)(2x+2)
and you just do all the algebra from there

[youngandinsane]

horizontal asymptote is anywhere where the slope=0
slope=rise over run so rise (or the y value) has to be equal to 0

0=(6x+24)(3x-1)/(8-x)(2x+2)
and you just do all the algebra from there

[youngandinsane]

horizontal asymptote is anywhere where the slope=0
slope=rise over run so rise (or the y value) has to be equal to 0

0=(6x+24)(3x-1)/(8-x)(2x+2)
and you just do all the algebra from there

[youngandinsane]

horizontal asymptote is anywhere where the slope=0
slope=rise over run so rise (or the y value) has to be equal to 0

0=(6x+24)(3x-1)/(8-x)(2x+2)
and you just do all the algebra from there

[youngandinsane]

horizontal asymptote is anywhere where the slope=0
slope=rise over run so rise (or the y value) has to be equal to 0

0=(6x+24)(3x-1)/(8-x)(2x+2)
and you just do all the algebra from there