Kirish

Kvadrat funksiya

y=x2−6x+5y=x^2-6x+5 funksiyani y=(x−3)2−4y=(x-3)^2-4 deb yozsak, hamma narsa ko'rinib qoladi: eng kichik qiymat −4-4, u x=3x=3 da olinadi, chapda funksiya kamayadi, o'ngda o'sadi. Imtihon masalalarining ko'pi shu uch sonni topishga keladi: parabola uchi, ildizlar va OyOy bilan kesishish nuqtasi. Bu dars parabolani uch shaklda yozishni va uning to'g'ri chiziq bilan munosabatini o'rgatadi.

Ta'rif va uch shakl

y=ax2+bx+cy=ax^2+bx+c, a≠0a\ne 0 ko'rinishidagi funksiya kvadrat funksiya deyiladi. Grafigi parabola. a=0a=0 bo'lsa, u kvadrat funksiya emas, chiziqli funksiyaga aylanadi.

ShaklYozuvDarhol ko'rinadi
ShaklUmumiyYozuvy=ax2+bx+cy=ax^2+bx+cDarhol ko'rinadiOyOy bilan kesishish: (0; c)(0;\,c)
ShaklUch orqaliYozuvy=a(x−m)2+ny=a(x-m)^2+nDarhol ko'rinadiuch: (m; n)(m;\,n)
ShaklIldizlar orqaliYozuvy=a(x−x1)(x−x2)y=a(x-x_1)(x-x_2)Darhol ko'rinadiOxOx bilan kesishish: x1, x2x_1,\ x_2

Masalan, y=x2−6x+5y=x^2-6x+5, y=(x−3)2−4y=(x-3)^2-4 va y=(x−1)(x−5)y=(x-1)(x-5) bir xil funksiyaning uch yozuvidir.

xy(3; −4)155
y=x2−6x+5y=x^2-6x+5: uchi (3; −4)(3;\,-4), ildizlari 11 va 55, OyOy bilan kesishish (0; 5)(0;\,5), simmetriya o'qi x=3x=3

Parabola uchi

ax2+bx+cax^2+bx+c
=a(x+b2a)2−D4a=a\left(x+\frac{b}{2a}\right)^2-\frac{D}{4a}
D=b2−4acD=b^2-4ac
To'liq kvadrat ajratish
x0=−b2ax_0=-\frac{b}{2a}
y0=f(x0)=−D4ay_0=f(x_0)=-\frac{D}{4a}
Parabola uchining koordinatalari

Amalda y0y_0 ni x0x_0 ni qo'yib topish osonroq. y=−3x2+12x−7y=-3x^2+12x-7 uchun x0=−122⋅(−3)=2x_0=-\dfrac{12}{2\cdot(-3)}=2, y0=−12+24−7=5y_0=-12+24-7=5. Formula bo'yicha ham: D=144−84=60D=144-84=60, −604⋅(−3)=5-\dfrac{60}{4\cdot(-3)}=5.

To'liq kvadrat qo'lda shunday ajratiladi.

2x2−12x+72x^2-12x+7
=2(x2−6x)+7=2(x^2-6x)+7
=2((x−3)2−9)+7=2\left((x-3)^2-9\right)+7
=2(x−3)2−11=2(x-3)^2-11
Uch (3; −11)(3;\,-11)

Koeffitsiyentlar nimani anglatadi

KoeffitsiyentGrafikda
a>0a>0shoxlar yuqoriga, uchi eng past nuqta
a<0a<0shoxlar pastga, uchi eng yuqori nuqta
∣a∣|a| kattaparabola tor, ∣a∣|a| kichik bo'lsa keng
ccOyOy bilan kesishish nuqtasi: f(0)=cf(0)=c
ba>0\dfrac{b}{a}>0uchi OyOy dan chapda, x0<0x_0<0
ba<0\dfrac{b}{a}<0uchi OyOy dan o'ngda, x0>0x_0>0
b=0b=0uchi OyOy da, funksiya juft
a+b+ca+b+cf(1)f(1) qiymati
a−b+ca-b+cf(−1)f(-1) qiymati
xy(1; 4)−133
y=−x2+2x+3y=-x^2+2x+3: a<0a<0, shoxlar pastga, uchi (1; 4)(1;\,4), eng katta qiymat 44

Xossalari

Xossaa>0a>0a<0a<0
XossaAniqlanish sohasia>0a>0(−∞; +∞)(-\infty;\,+\infty)a<0a<0(−∞; +∞)(-\infty;\,+\infty)
XossaQiymatlar sohasia>0a>0[y0; +∞)[y_0;\,+\infty)a<0a<0(−∞; y0](-\infty;\,y_0]
XossaKamayadia>0a>0(−∞; x0](-\infty;\,x_0]a<0a<0[x0; +∞)[x_0;\,+\infty)
XossaO'sadia>0a>0[x0; +∞)[x_0;\,+\infty)a<0a<0(−∞; x0](-\infty;\,x_0]
XossaUchidagi qiymata>0a>0eng kichika<0a<0eng katta

Parabola x=x0x=x_0 to'g'ri chiziqqa nisbatan simmetrik: f(x0+t)=f(x0−t)f(x_0+t)=f(x_0-t). Ikki nuqtaning ordinatalari teng bo'lsa, uch abssissasi ular o'rtasida. Xususan, ildizlar bo'lsa: x0=x1+x22x_0=\dfrac{x_1+x_2}{2}.

Shuning uchun nollarning o'rta arifmetigi uch abssissasiga teng. Uch (−2; −5)(-2;\,-5) bo'lsa va nollar mavjud bo'lsa, ularning o'rta arifmetigi −2-2.

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