Let f(x) =\frac{2x^2+2 x -40}{x-4}
Show that f(x) has a removable discontinuity at x=4 and determine what value for f(4) would make f(x) continuous at x=4.

Question

Let f(x) =\frac{2x^2+2 x -40}{x-4} 
Show that f(x) has a removable discontinuity at x=4 and determine what value for f(4) would make f(x) continuous at x=4.

kainui · Answer

So if you can factor the top part out and remove an x-4 term from the top, you'll have the exact same graph except with the hole filled in. Plug in 4, and you'll see exactly where it should be.

gorv · Answer

2x^2+2 x -40=2x2+10x-8x-40
2x(x+5)-8(x+5)
(x+5)(2x-8)
(x+5)*2*(x-4)

now denominator is (x-4)  so cancel out x-4
=2(x+5)
at x=4
=2*(4+5)=2*9=18  @abjectbrownE

anonymous · Answer

thanks man

nincompoop · Answer

you can make the equation continuous so long as your denominator will not be zero
you can achieve this by factoring the numerator so that a part of it matches your denominator