Testo Completa il quadrato, evidenziando le coordinate del vertice: (a) y=3+3x−5x2y=3+3x-5x^2y=3+3x−5x2 (b) y=2x23−x+1y=\dfrac{2x^2}{3}-x+1y=32x2−x+1 Soluzione (a) y=−5(x2−35x)+3=−5(x−310)2+5⋅9100+3=−5(x−310)2+6920y=-5\left(x^2-\tfrac{3}{5}x\right)+3=-5\left(x-\tfrac{3}{10}\right)^2+5\cdot\tfrac{9}{100}+3=-5\left(x-\tfrac{3}{10}\right)^2+\tfrac{69}{20}y=−5(x2−53x)+3=−5(x−103)2+5⋅1009+3=−5(x−103)2+2069. Vertice V(310;6920)=(0,3; 3,45)V\left(\tfrac{3}{10};\tfrac{69}{20}\right)=(0{,}3;\,3{,}45)V(103;2069)=(0,3;3,45). (b) y=23(x2−32x)+1=23(x−34)2−23⋅916+1=23(x−34)2+58y=\tfrac23\left(x^2-\tfrac32 x\right)+1=\tfrac23\left(x-\tfrac34\right)^2-\tfrac23\cdot\tfrac{9}{16}+1=\tfrac23\left(x-\tfrac34\right)^2+\tfrac58y=32(x2−23x)+1=32(x−43)2−32⋅169+1=32(x−43)2+85. Vertice V(34;58)=(0,75; 0,625)V\left(\tfrac34;\tfrac58\right)=(0{,}75;\,0{,}625)V(43;85)=(0,75;0,625). Va(310;6920),Vb(34;58)\boxed{V_a\left(\tfrac{3}{10};\tfrac{69}{20}\right),\qquad V_b\left(\tfrac34;\tfrac58\right)}Va(103;2069),Vb(43;85) Collegamenti Argomenti: Parabola Concetti: Completamento del quadrato · Forma vertice · Parabola · Vertice Funzioni: Parabola Competenze: Calcolare · Usare formule Tipo di esercizio: Calcolo espressioni