Question

Given the function: f(x) = 22-32+2 3x3 +1822 +27.0 1. Find the zeroes of this function. 2. Find the vertical asymptotes of this function. 3. Find the horizontal asymptote of this function. 4. Evaluate f(2)

Answer 1

1. 
   '10 +27% Given f(x)= x²-3x+2 373+18X2+27X find (1-) zeloes of the function - ie, set f(x)=0 72_34+2 3x3 + 18x2+278 72-37+2=0 X2-27-X+2 = 0 x(x-2) -1(2-2)=0 (x-2)(x-1) = 0 3-2=0, 3-15o =)x=1 =>x=2 . - 1, - о zel oes of - this function. ® vertical asymptotes of the function n=1 is a vertical asymptote where denominato isequal to f(x)=x²-3x+2 393+18x2+277 CS$canned with CamScanner
   
   denominato= 3x3 +18x²+274 = 3%C2% +6+9] Fr(4ty? - a²f6²tzabl E3*(*? + 2(3) x+(332] denominator = 3x (x+3)2 denominat or=o . 3x(x+3)2 =0 37=0 ,62+3) 2=0 )%+320 >)*=0 715-3 vertical - asymptotes are x=0, -37v Horizontal asymptotes: y=2 is a horizontal asymptore - if either lim f(2)=L (os) kno lim f(x)=L, and Lis finite. 28-00 1. x²-32+2 - = lim nuo 3x3+18 7²7274 nuo 33x x3 x3 3x3 x 3 + 3x3 1732720 Csscanned with CamScanner x3
   
   Dividing each term in numerato and denominat or by x3. = lim l / h - 2 2 2 2 2 2 2 2 ) *** ( 3+ + 2O aso t 18 27 18 & 72 3 toto so - 3+ 18 +27 22 Lilim *2=342 2400373718x2 +27% , aut x=-x tim u ²-3x+2 xs-oo 3x3 + 1821²+278 - lim 1-2)=36-X1+2 t o 3-x)3 +18(-x)+27(-4) lim x²+3x+2 -31²+18x²-270 usoo - تمه 18 -27 CS Scanned with CamScanner
   
   lim sro n²-3x+2 3234842+278 or " Horizontal asymptore is 1 y=10 ~ (4) To find of (2) 1. f(x) = x²-3x+2 3x3 + 18x12 +278 (2) ²-3 (2) +2 3(2)3+18 (2)2+2712) 4-6+ 2 -CO 150 24+72+54 if(2)=bv. CSScanned with CamScanner