Giá trị lớn nhất của hàm số \(y=-x^{2}-x+15\) là
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfKttLearuGlw5gvP1wzaeXatLxBI9gBam % XvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2DaeHbuLwB % Lnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFf % euY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9 % q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqaba % WaaqaafaaakqaabeqaaabaaaaaaaaapeGaemyEaKNaeyypa0JaeyOe % I0IaemiEaG3damaaCaaaleqabaWdbiabikdaYaaakiabgkHiTiabdI % ha4jabgUcaRiabigdaXiabiwda1aqaaiabg2da9iabgkHiTmaabmaa % baGaemiEaG3damaaCaaaleqabaWdbiabikdaYaaak8aacqGHRaWkpe % GaemiEaGNaeyOeI0IaeGymaeJaeGynaudacaGLOaGaayzkaaaabaGa % eyypa0JaeyOeI0YaaeWaaeaacqWG4baEdaahaaWcbeqaaiabikdaYa % aakiabgUcaRiabikdaYiabc6caUiabdIha4jabc6caUmaalaaabaGa % eGymaedabaGaeGOmaidaaiabgUcaRmaalaaabaGaeGymaedabaGaeG % inaqdaaiabgkHiTmaalaaabaGaeGymaedabaGaeGinaqdaaiabgkHi % TiabigdaXiabiwda1aGaayjkaiaawMcaaaqaaiabg2da9iabgkHiTm % aadmaabaWaaeWaaeaacqWG4baEcqGHRaWkdaWcaaqaaiabigdaXaqa % aiabikdaYaaaaiaawIcacaGLPaaadaahaaWcbeqaaiabikdaYaaaki % abgkHiTmaalaaabaGaeGOnayJaeGymaedabaGaeGinaqdaaaGaay5w % aiaaw2faaaqaaiabg2da9iabgkHiTmaabmaabaGaemiEaGNaey4kaS % YaaSaaaeaacqaIXaqmaeaacqaIYaGmaaaacaGLOaGaayzkaaWaaWba % aSqabeaacqaIYaGmaaGccqGHRaWkdaWcaaqaaiabiAda2iabigdaXa % qaaiabisda0aaaaaaa!834B! \begin{array}{l} y = - {x^2} - x + 15\\ = - \left( {{x^2} + x - 15} \right)\\ = - \left( {{x^2} + 2.x.\frac{1}{2} + \frac{1}{4} - \frac{1}{4} - 15} \right)\\ = - \left[ {{{\left( {x + \frac{1}{2}} \right)}^2} - \frac{{61}}{4}} \right]\\ = - {\left( {x + \frac{1}{2}} \right)^2} + \frac{{61}}{4} \end{array}\)
Ta có
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfKttLearuGlw5gvP1wzaeXatLxBI9gBam % XvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2DaeHbuLwB % Lnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFf % euY-Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0-yr0RYxir-Jbba9 % q8aq0-yq-He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqaba % Waaqaafaaakqaabeaeaaaaaaaaa8qabaGaeyOeI0YaaeWaaeaacqWG % 4baEcqGHRaWkdaWcaaqaaiabigdaXaqaaiabikdaYaaaaiaawIcaca % GLPaaadaahaaWcbeqaaiabikdaYaaakiabgsMiJkabicdaWaqaaiab % gkDiElabgkHiTmaabmaabaGaemiEaGNaey4kaSYaaSaaaeaacqaIXa % qmaeaacqaIYaGmaaaacaGLOaGaayzkaaWaaWbaaSqabeaacqaIYaGm % aaGccqGHRaWkdaWcaaqaaiabiAda2iabigdaXaqaaiabisda0aaacq % GHKjYOdaWcaaqaaiabiAda2iabigdaXaqaaiabisda0aaaaaaa!5872! \begin{array}{l} - {\left( {x + \frac{1}{2}} \right)^2} \le 0\\ \Rightarrow - {\left( {x + \frac{1}{2}} \right)^2} + \frac{{61}}{4} \le \frac{{61}}{4} \end{array}\)
Vậy giá trị lớn nhất của hàm số là \(\max y=\frac{61}{4}\)