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4.2-bo'lim : Parabolalar

6. Quyidagi parabolaning grafigini chizing. Grafik cho'qqi, \(y\)-kesish, \(x\)-kesish (agar mavjud bo'lsa) va cho'qqining har ikki tomonida kamida bitta nuqtadan iborat bo'lishi kerak.

\[f\left( x \right) = 4{x^2} - 4x + 1\]
Yechim

Avval cho'qqini topamiz. Bu holda tenglama \(f\left( x \right) = a{x^2} + bx + c\) shaklida bo'ladi. Shunday qilib, biz cho'qqi \(\left( { - \frac{b}{{2a}},f\left( { - \frac{b}{{2a}}} \right)} \right)\) nuqtasi ekanligini bilamiz. U holda cho'qqi,

\[\left( { - \frac{{ - 4}}{{2\left( 4 \right)}},f\left( { - \frac{{ - 4}}{{2\left( 4 \right)}}} \right)} \right) = \left( {\frac{1}{2},f\left( {\frac{1}{2}} \right)} \right) = \left( {\frac{1}{2},0} \right)\]

Shuni ham yodda tutingki, bu parabola uchun \(a = 4 > 0\) va shuning uchun parabola yuqoriga ochiladi.

2-qadam

\(y\)-kesishma faqat \(\left( {0,f\left( 0 \right)} \right)\) nuqtasidir. Funktsiyani tezkor baholash bizga \(f\left( 0 \right) = 1\) va shunga o'xshash tenglamamiz uchun \(y\)-kesishma \(\left( {0,1} \right)\) ekanligini beradi.

3-qadam

\(x\)-kesishmalar uchun biz odatda \(f\left( x \right) = 0\) tenglamasini yechamiz. Biroq, bu holatda biz buni qilishimiz shart emas. Birinchi qadamdan biz cho'qqining a borligini ko'ramiz y‑nol koordinatasi va shuning uchun ham \(x\)-kesishma hisoblanadi. Bundan tashqari, u cho'qqi bo'lgani uchun bu funktsiya uchun yagona \(x\)-kesishmasi bo'lishi mumkin.

E'tibor bering, agar biz tenglamani yechganimizda, biz ham mana shu yagona \(x\)-kesishmasiga kelgan bo'lardik.

4-qadam

Bu holda bizda faqat cho'qqi (yagona \(x\)-kesishma bo'ladi) va \(y\)-kesishma (cho'qqining chap tomonida joylashgan) mavjud. Shunday qilib, bizga cho'qqining chap tomonida joylashgan nuqta kerak bo'ladi va biz bu nuqta uchun \(y\)-kesishmasiga mos keladigan cho'qqining chap tomonidagi nuqtani topishimiz mumkin.

\(y\)-kesish cho'qqisining chap tomonidagi 1 masofadir va shuning uchun o'ng tomonda bir xil \(y\) qiymatida mos keladigan nuqta bo'ladi va u cho'qqining o'ng tomonida 1 masofa bo'ladi. Shuning uchun, ga mos keladigan cho'qqining o'ng tomonidagi nuqta y‑tutash \(\left( {1,1} \right)\).

5-qadam

Mana, biz yuqorida topilgan barcha nuqtalarni o'z ichiga olgan parabolaning eskizi.

The graph of the parabola \(f\left( x \right) = 4{x^2} - 4x + 1\).  It opens upward and its vertex just touches the x-axis so there is a single x-intercept.  The points found above are marked and labeled on it and they are the vertex, which is also the only x-intercept, \(\left( {\frac{1}{2},0} \right)\), the y-intercept \(\left( {0,1} \right)\) and the extra point \(\left( {1,1} \right)\).