Odpowiedź :
Wzory:
[tex]\Delta=b^2-4ac\\x_1=\dfrac{-b-\sqrt{\Delta}}{2a}\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}\\p=\dfrac{-b}{2a}\\q=\dfrac{-\Delta}{4a}[/tex]
Obliczenia:
[tex]f(x)=-3x^2-5x+2[/tex]
[tex]\Delta=(-5)^2-4\cdot(-3)\cdot2\\\Delta=25+24\\\Delta=49\\\sqrt{\Delta}=7[/tex]
[tex]x_1=\dfrac{-(-5)-7}{2\cdot(-3)}=\dfrac{5-7}{-6}=\dfrac{-2}{-6}=\dfrac{1}{3}[/tex]
[tex]x_2=\dfrac{-(-5)+7}{2\cdot(-3)}=\dfrac{5+7}{-6}=\dfrac{12}{-6}=-2[/tex]
[tex]p=\dfrac{-(-5)}{2\cdot(-3)}=\dfrac{5}{-6}=-\dfrac{5}{6}[/tex]
[tex]q=\dfrac{-49}{4\cdot(-3)}=\dfrac{-49}{-12}=\dfrac{49}{12}[/tex]
Postać iloczynowa:
[tex]f(x)=a(x-x_1)(x-x_2)\\f(x)=-3(x-\dfrac{1}{3})(x-(-2))\\f(x)=-3(x-\dfrac{1}{3})(x+2)[/tex]
Postać kanoniczna:
[tex]f(x)=a(x-p)^2+q\\f(x)=-3\left(x-\left(-\dfrac{5}{6}\right)\right)^2+\dfrac{49}{12}\\f(x)=-3\left(x+\dfrac{5}{6}\right)^2+\dfrac{49}{12}[/tex]
Postać ogólna funkcji kwadratowej:
f(x) = ax² + bx + c
f(x) = -3x² - 5x + 2
a = -3, b = -5, c = 2
Postać kanoniczna funkcji kwadratowej to:
f(x) = a(x - p)² + q
gdzie: [tex]\bold{p=\dfrac{-b}{2a}\,,\quad q=\dfrac{-\Delta}{4a}\,,\quad \Delta=b^2-4ac}[/tex]
Stąd:
[tex]\bold{p=\dfrac{-(-5)}{2(-3)}=\dfrac56}\\\\\bold{ \Delta=(-5)^2-4(-3)\cdot2=25+24=49} \\\\\bold{ q=\dfrac{-49}{4(-3)}=\dfrac{49}{12}}\\\\ \large\boxed{\bold{f(x)=-3\left(x-\frac56\right)^2+\frac{49}{12}}}[/tex]
Postać iloczynowa funkcji kwadratowej to:
f(x) = a(x - x₁)(x - x₂)
gdzie: [tex]\bold{x_1=\dfrac{-b-\sqrt\Delta}{2a}\,,\quad x_2=\dfrac{-b+\sqrt\Delta}{2a}}[/tex]
Stąd:
[tex]\bold{\Delta=49\quad\implies\quad \sqrt\Delta=7}\\\\\bold{x_1=\dfrac{-(-5)-7}{2(-3)}=\dfrac2{-6}=-\dfrac13\,,\quad x_2=\dfrac{-(-5)+7}{2(-3)}=\dfrac{12}{-6}=-2} \\\\\\\large\boxed{\bold{f(x)=-3\left(x+\frac13\right)\big(x+2\bigt)}}[/tex]