Cho hàm số bậc ba \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadggacaWG4bWaaWba % aSqabeaacaaIZaaaaOGaey4kaSIaamOyaiaadIhadaahaaWcbeqaai % aaikdaaaGccqGHRaWkcaWGJbGaamiEaiabgUcaRiaadsgaaaa!458D! f\left( x \right) = a{x^3} + b{x^2} + cx + d\) có đồ thị như hình vẽ bên. Hỏi đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaWaaeWaaeaa % caWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaG4maiaadIhacq % GHRaWkcaaIYaaacaGLOaGaayzkaaWaaOaaaeaacaaIYaGaamiEaiab % gUcaRiaaigdaaSqabaaakeaadaqadaqaaiaadIhadaahaaWcbeqaai % aaisdaaaGccqGHsislcaaI1aGaamiEamaaCaaaleqabaGaaGOmaaaa % kiabgUcaRiaaisdaaiaawIcacaGLPaaacaGGUaGaamOzamaabmaaba % GaamiEaaGaayjkaiaawMcaaaaaaaa!528F! g\left( x \right) = \frac{{\left( {{x^2} - 3x + 2} \right)\sqrt {2x + 1} }}{{\left( {{x^4} - 5{x^2} + 4} \right).f\left( x \right)}}\) có bao nhiêu đường tiệm cận đứng?

Quan sát đồ thị hàm số f(x) ta thấy đồ thị hàm số cắt trục hoành tại điểm có hoành độ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaaIWaaabeaakabaaaaaaaaapeGaeyicI48aaeWaaeaacaaI % WaGaai4oaiaaigdaaiaawIcacaGLPaaaaaa!3D42! {x_0} \in \left( {0;1} \right)\), có hệ số a > 0 và tiếp xúc với trục hoành tại điểm có hoành độ bằng 2. Từ đó suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadggadaqadaqaaiaa % dIhacqGHsislcaWG4bWaaSbaaSqaaiaaicdaaeqaaaGccaGLOaGaay % zkaaWaaeWaaeaacaWG4bGaeyOeI0IaaGOmaaGaayjkaiaawMcaamaa % CaaaleqabaGaaGOmaaaaaaa!45C9! f\left( x \right) = a\left( {x - {x_0}} \right){\left( {x - 2} \right)^2}\).

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaWaaeWaaeaa % caWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaG4maiaadIhacq % GHRaWkcaaIYaaacaGLOaGaayzkaaWaaOaaaeaacaaIYaGaamiEaiab % gUcaRiaaigdaaSqabaaakeaadaqadaqaaiaadIhadaahaaWcbeqaai % aaisdaaaGccqGHsislcaaI1aGaamiEamaaCaaaleqabaGaaGOmaaaa % kiabgUcaRiaaisdaaiaawIcacaGLPaaacaGGUaGaamOzamaabmaaba % GaamiEaaGaayjkaiaawMcaaaaacqGH9aqpdaWcaaqaamaabmaabaGa % amiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiodacaWG4bGaey % 4kaSIaaGOmaaGaayjkaiaawMcaamaakaaabaGaaGOmaiaadIhacqGH % RaWkcaaIXaaaleqaaaGcbaWaaeWaaeaacaWG4bWaaWbaaSqabeaaca % aI0aaaaOGaeyOeI0IaaGynaiaadIhadaahaaWcbeqaaiaaikdaaaGc % cqGHRaWkcaaI0aaacaGLOaGaayzkaaGaaiOlaiaadggadaqadaqaai % aadIhacqGHsislcaWG4bWaaSbaaSqaaiaaicdaaeqaaaGccaGLOaGa % ayzkaaWaaeWaaeaacaWG4bGaeyOeI0IaaGOmaaGaayjkaiaawMcaam % aaCaaaleqabaGaaGOmaaaaaaaaaa!73A5! g\left( x \right) = \frac{{\left( {{x^2} - 3x + 2} \right)\sqrt {2x + 1} }}{{\left( {{x^4} - 5{x^2} + 4} \right).f\left( x \right)}} = \frac{{\left( {{x^2} - 3x + 2} \right)\sqrt {2x + 1} }}{{\left( {{x^4} - 5{x^2} + 4} \right).a\left( {x - {x_0}} \right){{\left( {x - 2} \right)}^2}}}\) xác định trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9maajibabaGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaacaGG % 7aGaey4kaSIaeyOhIukacaGLBbGaayzkaaGaaiixamaacmaabaGaam % iEamaaBaaaleaacaaIWaaabeaakiaacYcacaaIXaGaaiilaiaaikda % aiaawUhacaGL9baaaaa!46F1! D = \left[ { - \frac{1}{2}; + \infty } \right)\backslash \left\{ {{x_0},1,2} \right\}\) và  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaWaaOaaaeaa % caaIYaGaamiEaiabgUcaRiaaigdaaSqabaaakeaacaWGHbWaaeWaae % aacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaamaabmaabaGaamiE % aiabgUcaRiaaikdaaiaawIcacaGLPaaadaqadaqaaiaadIhacqGHsi % slcaaIYaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOWaaeWa % aeaacaWG4bGaeyOeI0IaamiEamaaBaaaleaacaaIWaaabeaaaOGaay % jkaiaawMcaaaaaaaa!51A6! g\left( x \right) = \frac{{\sqrt {2x + 1} }}{{a\left( {x + 1} \right)\left( {x + 2} \right){{\left( {x - 2} \right)}^2}\left( {x - {x_0}} \right)}}\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcaWG4bWaaSbaaWqa % aiaaicdaaeqaaSWaaWbaaWqabeaacqGHRaWkcaGGVaGaeyOeI0caaa % WcbeaakiaadEgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqp % qaaaaaaaaaWdbiabgglaX+aacqGHEisPcaGGSaWaaCbeaeaaciGGSb % GaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcaaIYaWaaWbaaWqabeaa % cqGHRaWkcaGGVaGaeyOeI0caaaWcbeaakiaadEgadaqadaqaaiaadI % haaiaawIcacaGLPaaacqGH9aqppeGaeyySae7daiabg6HiLcaa!5ABB! \mathop {\lim }\limits_{x \to {x_0}^{ + / - }} g\left( x \right) = \pm \infty ,\mathop {\lim }\limits_{x \to {2^{ + / - }}} g\left( x \right) = \pm \infty \) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcaaIXaaabeaakiaa % dEgacaGGOaGaamiEaiaacMcaaaa!3FEE! \mathop {\lim }\limits_{x \to 1} g(x)\) hữu hạn nên hàm số có 2 tiệm cận đứng là \(x = x_0\) và x = 2.