Did I do this right? And how do I know if there is holes? f(x) =0=-3x (x+2)FEATURES:(Write "none" where appropriate.)Equations of all vertical asymptotes:x=1(x,y)-coordinates of all intercepts:(0,0), (-2,0)(0,0)-3.0(01+2)о2((0)-1)²2 (0-1)/² = 1 = 0(x,y)-coordinates of all "holes":Equation for horizontal asymptote:y= -3/2-3x(x + 2)2(x - 1)²-3×(x+2) = 0Equation for slant asymptote:No slant asymptote-3x² - 6x2x² - 4x + 2x+2=0x20-2 -2 X2-2x=AAJAL

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Answered: f(x) = -3x² - 6x 2x² - 4x + 2 = -3x(x +…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.5: Rational Functions

Problem 61E

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f(x) = -3x² - 6x 2x² - 4x + 2 = -3x(x + 2) 2(x - 1)²