Indlela yokufumana phezulu parabola, ndibuye ndiyakhe

Kwimathematika, kukho uthotho yonke kobubani, phakathi apho indawo ebalulekileyo yile quadratic equation. ukulingana zinokuqwalaselwa zombini ngokwahlukeneyo kwaye ekunikeni phezu nokulawula nezamazembe. Iingcambu square iikoteshini ezi ndawo engenelayo parabola kunye ngqo oh.

umbono jikelele

Quadratic equation sisonke isakhiwo kulandelayo:

kugawulwa ² + 'bx' + c = 0

Xa indima "X kaThixo" baphathwa njengabantu izinto ezahlukeneyo, kwaye lonke nkulumo. Umzekelo:

2x ² + 5x-4 = 0;

(X + 7) ² +3 (x + 7) + 2 = 0.

Kwimeko apho x limi njengoko njengembonakaliso, kuyimfuneko ukuba isondezwe ibe variable kwaye ufumane iingcambu lenxaki. Emva koko, ukuba umyinge polynomial nokusombulula ye x.

Ngoko ke, ukuba (x + 7) = a, inxaki kuthatha uhlobo ² + 3a + 2 = 0.

A = ^{3 2 -4 * 1 * 2 =} 1 ;

kunye ₁ = (- 3-1) / 2 * 1 = -2;

₂ = (- 3 + 1) / 2 * 1 = -1 .

Xa iingcambu olulinganayo -1 no -2, siye sifumane ezi zilandelayo:

x + 7 = 2 x + 7 = -1;

x = -9 kunye x = -8.

Iingcambu ke amaxabiso x-zilungelelanise lwendawo kweendlela kunye abscissa le parabola. Enyanisweni, ukubaluleka kwabo ayibalulekanga xa injongo kuphela ukufumana encotsheni parabola. Kodwa ukuhlela iingcambu zidlala indima ebalulekileyo.

Indlela yokufumana encotsheni parabola

Makhe siye umva equation yokuqala. Ukuze uphendule umbuzo yokufumana njani encotsheni parabola, kuyimfuneko ukwazi le fomyula ilandelayo:

x _sn = -b / 2a,

apho x _sn - ixabiso lwe x-Inanai elihambiselanayo lika kwinqanaba elifunekayo.

Kodwa ukufumana njani encotsheni parabola ngaphandle ixabiso y-isilungelelanisi? We shintsha ixabiso ifunyenwe lenxaki x kwaye ufumane variable oyifunayo. Ngokomzekelo, siya ukusombulula inxaki kulandelayo:

x ² + 3 = 5 0

Thina ukufumana ixabiso le x-zilungelelanise ukuba enekona le parabola:

x _sn = -b / 2a = -3 / 2 * 1;

x _sn = -1,5.

Fumana ixabiso y-zilungelelanise ukuba enekona le parabola:

y = 2x ² + '4x 3 = (- 1.5) ² +3 * (- 1,5) -5;

y = -7,25.

Isiphumo kukuba kwincopho parabola libekwe ku zilungelelanise (-1,5; -7.25).

Ukwakhiwa parabola

A parabola yikhompawundi iingongoma ukuba nkqo ongu-macala. Ngenxa yesi sizathu, ukwakhiwa kwawo ayikho nzima. Eyona nto kunzima - kukuba ukwenza izibalo echanekileyo zilungelelanise zamanqaku.

Kufuneka sinikele ingqalelo ngokukhethekileyo okuza lenxaki equation.

Ngumlingani kuchaphazela ulwalathiso parabola. Kwimeko xa sele esinentlawulo engenaxabiso, amasebe amele phantsi, lo uphawu oluvumelanayo - up.

Lomlingani b ubonisa ububanzi i parabola ngesandla. I elikhulu ixabiso, kokukhona kuya kuba.

Le lomlingani kubonisa ukufuduka embindini ongu y kuthelekiswa ngemvelaphi parabola.

Indlela yokufumana encotsheni parabola, sele sifundile, nokuba ukufumana iingcambu, kufuneka bakhokelwe zezi zinto zilandelayo:

D = b ² -4ac,

apho D - na discriminant, oyimfuneko ukwenzela ukufumana esineengcambu alenxaki.

x ₁ = ^(- b + V - D) / 2a

x ₂ = ^(- BV - D) / 2a

Iinqobo ze x uya kunxibelelana ukugxila amaxabiso ze y, njengoko Bona iingongoma yonqumlo kunye x-axis.

Emva koko siphawula kwi ukuqhuba moya le enekona le parabola kunye amaxabiso ifunyenwe. Ukuze ishedyuli eneenkcukacha kuyimfuneko ukufumana iingongoma ezimbalwa kakhulu. Ukuza kuthi ga ngoku khetha nasiphi na ixabiso x, idomeyini kuvumelekile, kwaye endaweni kwi Isibalo function. Isiphumo izibalo nguye Inanai elihambiselanayo lika kwindawo embindini ongu y.

Lula le nkqubo yokwakha ucwangciso, ungenza ukrwele umgca nkqo ngokusebenzisa enekona le parabola kwaye aa Incopho ohamba embindini ongu x. Oku kuya kuba esqwini wolingano- macala, esebenzisa apho, ukuba ingongoma enye, singachazwa kunye equidistant yesibini ukusuka kumgca azotywe.