Question

9x2+22x+9 x3+5x2+7x+3 a. Factor the numerator and denominator using the rational zero theorem as needed). Include any factors with irrational zeros but leave quadratic irreducibles as they are (no real zeros) b. State all zeros of f(x) (excluding any complex). c. Determine all asympotes (lines) and any holes (points) if they exist.. d. Write all intercepts as points. Plot all the points and asympotes so far by drawing your own appropriate scale using a ruler. e. Use extra points as needed to describe the shape between and after zeros. You may want to do this before picking your scale to make sure the y-axis fits. f. State in interval notation when f(x) 2 0 (make sure and use union to write as a complete answer and set brackets for any isolates zeros. 2. f(x) =

Answer 1

☆ y = gx² + 22x + 9 x3 + x² +72+3. here numerator is irreducible. denominator is, x3 + x² + 7x + 3 = 0 here ao 3, an=1 dividers of an: 1,3 dividers of ao. 1 rational members: I 1,3 so take here - synthetic -1 i lo I to is a root division S 7 -I aty 4 3 to z 3 -3 0 x +48 +3=0 ax² + 3x + x + 3=0 al ( +3 ) tila +3220 CO+ 32 (2+1)=0

y = gx² +222 + 9 (a +1] (ac+3) Colti) 14 = 9x² +222 + 9 kcal. (2+112 (x+3) to find zeros, take yo gac² +22a + g =0 (x+112 (2+3] 9x² +222 +9=0 x= - b ± √ 62-496 29 209) x = -22_+J222-469269) ME -22 3 1484 - 324 18

x = – 22 I 1160 18 x-intercepts are ! (-** 2x10,0) and (-11-2 JO ,u) * to find ventical asymptote, take denominator =0 (€ +1] 2 (2+3) = 0 x+1=0, 2+3=0 vertical asymptote

here degree of numerator is greater than the degree of denominator. so horizontal asymptote is x-axis ire yo so horizontal asymptote is 1450 take x=0 * to find y-intercepts = 9(0)² + 22 (0) +9 3 + sco)² +7(0) +3 Y 0 +0 +9 0+0+0+3 G=3 y-intercept: (0,3) x-intencepts we have already find. ("taso, o ) and (-"- 2010, 0)

☆ to x few points graph the line weneed 1 y = f(x2 -2.66 2-5 1.978 1.625 fead so on from the graph - interval. 103, -52[co] •[102JO , co)

f(x)>=0 -15