Przedstaw funkcję kwadratową w postaci ogólnej

y=3(x-3)2+1

y=-2(x+1)2

y=-(x-1)2-2

y= 3(x-10)2-6

y=8(x+5)2+1



Odpowiedź :

Cześć!

Postać ogólna funkcji kwadratowej

[tex]y=ax^2+bx+c[/tex]

1)

[tex]y=3(x-3)^2+1\\\\y=3(x^2-2\cdot x\cdot3+3^2)+1\\\\y=3(x^2-6x+9)+1\\\\y=3\cdot x^2-3\cdot6x+3\cdot9+1\\\\y=3x^2-18x+27+1\\\\\huge\boxed{y=3x^2-18x+28}[/tex]

2)

[tex]y=-2(x+1)^2\\\\y=-2(x^2+2\cdot x\cdot 1+1^2)\\\\y=-2(x^2+2x+1)\\\\y=-2\cdot x^2-2\cdot2x-2\cdot1\\\\\huge\boxed{y=-2x^2-4x-2}[/tex]

3)

[tex]y=-(x-1)^2-2\\\\y=-(x^2-2\cdot x\cdot1+1^2)-2\\\\y=-(x^2-2x+1)-2\\\\y=-x^2+2x-1-2\\\\\huge\boxed{y=-x^2+2x-3}[/tex]

4)

[tex]y=3(x-10)^2-6\\\\y=3(x^2-2\cdot x\cdot10+10)^2-6\\\\y=3(x^2-20x+100)-6\\\\y=3\cdot x^2-3\cdot20x+3\cdot100-6\\\\y=3x^2-60x+300-6\\\\\huge\boxed{y=3x^2-60x+294}[/tex]

5)

[tex]y=8(x+5)^2+1\\\\y=8(x^2+2\cdot x\cdot5+5^2)+1\\\\y=8(x^2+10x+25)+1\\\\y=8\cdot x^2+8\cdot10x+8\cdot25+1\\\\y=8x^2+80x+200+1\\\\\huge\boxed{y=8x^2+80x+201}[/tex]