Asosiy tarkibga o'tish

4.7-bo'lim: Simmetriya

Ushbu bo'limda biz parabolalarning grafigini chizganimizda ishlatgan narsani ko'rib chiqamiz. Biroq, biz ushbu bo'limda uning umumiy ko'rinishini ko'rib chiqamiz. Ko'p grafiklar mavjud simmetriya ularga.

Simmetriya tenglamaning grafigini tuzishda foydali bo'lishi mumkin, chunki agar biz grafikning bir qismini bilsak, grafikning qolgan (va simmetrik) qismini ham bilib olamiz. Biz ushbu faktdan ba'zi grafiklarning qo'shimcha nuqtasini olish uchun parabolalarning grafiklarini chizganimizda foydalandik.

Ushbu bo'limda biz simmetriyaning uchta turini ko'rib chiqmoqchimiz.

  1. Grafik deb aytiladi \(x\) o'qiga nisbatan simmetrik agar har doim \(\left( {a,b} \right)\) grafikda bo'lsa, \(\left( {a, - b} \right)\) ham shunday bo'ladi. Mana \(x\) o'qiga nisbatan simmetrik bo'lgan grafikning eskizi.
    There are no tick marks on the x or y-axis in this graph and the 1st and 4th quadrants only are shown. The function graphed looks to be a parabola with a vertex at the origin and it opens off to the right into the 1st and 4th quadrants. In the portion of the graph in the 1st quadrant a point is marked on the graph with coordinates (a,b). The portion of the graph in the 4th quadrant has a point marked in it with coordinates (a,-b). This is to illustrate that the portion of the graph in the 4th quadrant is basically a reflection of the portion in the 1st quadrant and for any x value, say x=a, there will be one point above it at y=b and a point below it at y=-b.
  2. Grafik deb aytiladi \(y\) o'qiga nisbatan simmetrik agar har doim \(\left( {a,b} \right)\) grafikda bo'lsa, \(\left( { - a,b} \right)\) ham shunday bo'ladi. Mana \(y\) o'qiga nisbatan simmetrik bo'lgan grafikning eskizi.
    There are no tick marks on the x or y-axis in this graph and the 1st and 2nd quadrants only are shown. The function graphed looks to be a parabola with a vertex at the origin and it opens off up into the 1st and 2nd quadrants. In the portion of the graph in the 1st quadrant a point is marked on the graph with coordinates (a,b). The portion of the graph in the 2nd quadrant has a point marked in it with coordinates (-a,b). This is to illustrate that the portion of the graph in the 2nd quadrant is basically a reflection of the portion in the 1st quadrant and for any y value, say y=b, there will be one point to the right at x=a and another point to the left at x=-a.
  3. Grafik deb aytiladi kelib chiqishiga nisbatan simmetrik agar har doim \(\left( {a,b} \right)\) grafikda bo'lsa, \(\left( { - a, - b} \right)\) ham shunday bo'ladi. Bu erda kelib chiqishi bo'yicha simmetrik bo'lgan grafikning eskizi.
    In the 1st quadrant there is a graph that contains the point (a,b) and as we increase x away from this point the graph slopes down towards the x-axis and flattens out as it gets closer to the x-axis but never crosses the x-axis. As we decrease x from this point towards the y-axis the graph increases rapidly getting closer and closer to the y-axis becoming almost vertical near the y-axis but it never crosses the y-axis. In the 3rd quadrant there is a graph that contains the point (-a,-b) and as we increase x away from this point in the negative direction the graph slopes up towards the x-axis and flattens out as it gets closer to the x-axis but never crosses the x-axis. As we decrease x from this point towards the y-axis the graph decreases rapidly getting closer and closer to the y-axis becoming almost vertical near the y-axis but it never crosses the y-axis. The point of this graph is to show that for any point in the 1st quadrant there will be a point in the 3rd quadrant with the same coordinates except opposite signs.

E'tibor bering, ko'pgina grafikalar hech qanday simmetriyaga ega emas. Bundan tashqari, grafik simmetriyaning bir nechta turlariga ega bo'lishi mumkin. Masalan, koordinata boshida joylashgan aylananing grafigi barcha uchta simmetriyani ko'rsatadi.

Simmetriya uchun testlar

Bizda simmetriyaning har xil turlari uchun juda oddiy sinovlar mavjud.

  1. Grafik \(x\) o'qiga nisbatan simmetriyaga ega bo'ladi, agar barcha \(y\) lar -\(y\) bilan almashtirilganda ekvivalent tenglamaga ega bo'lsak.

  2. Agar barcha \(x\) lar –\(x\) bilan almashtirilganda, ekvivalent tenglamaga ega bo‘lsak, grafik \(y\) o‘qiga nisbatan simmetriyaga ega bo‘ladi.

  3. Agar barcha \(y\) lar –\(y\) bilan almashtirilsa va barcha \(x\) lar –\(x\) bilan almashtirilsa, ekvivalent tenglamani oladigan bo‘lsak, grafik kelib chiqishi bo‘yicha simmetriyaga ega bo‘ladi.

Biz bunga misol keltirganimizda “ekvivalent tenglama” deganda nimani nazarda tutayotganimizni aniqlaymiz. Biz ko'rib chiqishimiz mumkin bo'lgan ko'pgina misollar uchun bu aynan bir xil tenglama ekanligini anglatadi.

Simmetriya uchun bir nechta tenglamalarni sinab ko'raylik. E'tibor bering, biz ularni grafika bilan ta'minlamaymiz, chunki ularning ko'pchiligini grafik qilish juda qiyin. Ushbu misolning maqsadi faqat har bir tenglamaning simmetriyasini aniqlash uchun testlardan foydalanishdir.

1-misol Quyidagi tenglamalarning har birining simmetriyasini aniqlang.
  1. \(y = {x^2} - 6{x^4} + 2\)
  2. \(y = 2{x^3} - {x^5}\)
  3. \({y^4} + {x^3} - 5x = 0\)
  4. \(y = {x^3} + {x^2} + x + 1\)
  5. \({x^2} + {y^2} = 1\)
a \(y = {x^2} - 6{x^4} + 2\) Yechim

Biz avval \(x\) o'qida simmetriyani tekshiramiz. Bu shuni anglatadiki, biz barcha \(y\)-larni –\(y\) bilan almashtirishimiz kerak. Buni qilish juda oson, chunki faqat bitta \(y\) mavjud.

\[ - y = {x^2} - 6{x^4} + 2\]

Endi bu ekvivalent tenglama emas, chunki o'ngdagi shartlar asl tenglama bilan bir xil, chapdagi atama esa qarama-qarshi belgidir. Shunday qilib, bu tenglama \(x\) o'qiga nisbatan simmetriyaga ega emas.

Keyinchalik, \(y\) o'qi bo'yicha simmetriyani tekshiramiz. Bu erda biz barcha \(x\)-larni –\(x\) bilan almashtiramiz.

\[\begin{align*}y & = {\left( { - x} \right)^2} - 6{\left( { - x} \right)^4} + 2\\ y & = {x^2} - 6{x^4} + 2\end{align*}\]

Soddalashtirgandan so'ng, biz aynan bir xil tenglamani qaytarib oldik, bu ikkalasi teng ekanligini anglatadi. Shuning uchun bu tenglama \(y\) o'qiga nisbatan simmetriyaga ega.

Nihoyat, biz kelib chiqishi bo'yicha simmetriyani tekshirishimiz kerak. Bu erda ikkala o'zgaruvchini almashtiramiz.

\[\begin{align*} - y & = {\left( { - x} \right)^2} - 6{\left( { - x} \right)^4} + 2\\ - y & = {x^2} - 6{x^4} + 2\end{align*}\]

Shunday qilib, birinchi sinovda bo'lgani kabi, chap tomon asl tenglamadan farq qiladi va o'ng tomon asl tenglama bilan bir xil. Shuning uchun, bu asl tenglamaga teng emas va bizda kelib chiqishi bo'yicha simmetriya yo'q.


b \(y = 2{x^3} - {x^5}\) Yechim

Biz bu erda juda ko'p tafsilotlarni keltirmaymiz. Birinchidan, biz \(x\) o'qi bo'yicha simmetriyani tekshiramiz.

\[ - y = 2{x^3} - {x^5}\]

Bu erda bizda simmetriya yo'q, chunki bir tomoni asl tenglama bilan bir xil, ikkinchisi esa yo'q. Shunday qilib, bizda \(x\) o'qida simmetriya yo'q.

Keyin \(y\) o'qining simmetriyasini tekshiring.

\[\begin{align*}y &= 2{\left( { - x} \right)^3} - {\left( { - x} \right)^5}\\ y &= - 2{x^3} + {x^5}\end{align*}\]

Esda tutingki, agar biz g'alati kuchga salbiy qabul qilsak, minus belgisi oldinga chiqishi mumkin. Shunday qilib, soddalashtirgandan so'ng, biz chap tomonni asl tenglamaga o'xshash qilib olamiz, lekin o'ng tomon endi asl tenglamaning qarama-qarshi belgisidir va shuning uchun bu asl tenglamaga ekvivalent emas va shuning uchun bizda \(y\) o'qida simmetriya yo'q.

Nihoyat, kelib chiqishi bo'yicha simmetriyani tekshiramiz.

\[\begin{align*} - y &= 2{\left( { - x} \right)^3} - {\left( { - x} \right)^5}\\ - y & = - 2{x^3} + {x^5}\end{align*}\]

Endi, bu safar ushbu tenglamadagi barcha belgilar asl tenglamaning mutlaqo teskarisi ekanligiga e'tibor bering. Bu shuni anglatadiki, u asl tenglamaga ekvivalentdir, chunki biz qilishimiz kerak bo'lgan yagona narsa dastlabki tenglamaga qaytish uchun hamma narsani "-1" ga ko'paytirishdir.

Shuning uchun, bu holda biz kelib chiqishi bo'yicha simmetriyaga egamiz.


c \({y^4} + {x^3} - 5x = 0\) Yechim

Birinchidan, \(x\) o'qining simmetriyasini tekshiring.

\[\begin{align*}{\left( { - y} \right)^4} + {x^3} - 5x& = 0\\ {y^4} + {x^3} - 5x &= 0\end{align*}\]

Bu asl tenglama bilan bir xil va shuning uchun biz \(x\) o'qida simmetriyaga egamiz.

Endi \(y\) o'qining simmetriyasini tekshiring.

\[\begin{align*}{y^4} + {\left( { - x} \right)^3} - 5\left( { - x} \right) &= 0\\ {y^4} - {x^3} + 5x &= 0\end{align*}\]

Shunday qilib, ba'zi atamalar asl tenglama bilan bir xil belgiga ega, boshqalari esa yo'q, shuning uchun \(y\) o'qida simmetriya yo'q.

Nihoyat, kelib chiqishi bo'yicha simmetriyani tekshiring.

\[\begin{align*}{\left( { - y} \right)^4} + {\left( { - x} \right)^3} - 5\left( { - x} \right) &= 0\\ {y^4} - {x^3} + 5x& = 0\end{align*}\]

Shunga qaramay, bu asl tenglama bilan bir xil emas va asl tenglamadan mutlaqo teskari belgi emas va shuning uchun kelib chiqishi nosimmetrik emas.


d \(y = {x^3} + {x^2} + x + 1\) Yechim

Birinchidan, \(x\) o'qi bo'yicha simmetriya.

\[ - y = {x^3} + {x^2} + x + 1\]

Bu \(x\) o'qida simmetriyaga o'xshamaydi

Keyinchalik, \(y\) o'qi bo'yicha simmetriya.

\[\begin{align*}y &= {\left( { - x} \right)^3} + {\left( { - x} \right)^2} + \left( { - x} \right) + 1\\ y &= - {x^3} + {x^2} - x + 1\end{align*}\]

Shunday qilib, bu erda ham simmetriya yo'q.

Nihoyat, kelib chiqishi bo'yicha simmetriya.

\[\begin{align*} - y &= {\left( { - x} \right)^3} + {\left( { - x} \right)^2} + \left( { - x} \right) + 1\\ - y& = - {x^3} + {x^2} - x + 1\end{align*}\]

Va yana, bu erda ham simmetriya yo'q.

Bu funksiya hech qanday simmetriyaga ega emas. Bu g'alati emas, chunki ko'pchilik funktsiyalarda bu simmetriyalarning hech biri yo'q.


e \({x^2} + {y^2} = 1\) Yechim

Avval \(x\) o'qi simmetriyasini tekshiring.

\[\begin{align*}{x^2} + {\left( { - y} \right)^2} &= 1\\ {x^2} + {y^2} &= 1\end{align*}\]

Shunday qilib, u \(x\) o'qi simmetriyasiga nisbatan simmetriyaga ega.

Keyin \(y\) o'qi simmetriyasini tekshiring.

\[\begin{align*}{\left( { - x} \right)^2} + {y^2} &= 1\\ {x^2} + {y^2} &= 1\end{align*}\]

Aftidan, u \(y\) o'qi simmetriyasiga ham ega.

Nihoyat, kelib chiqishi bo'yicha simmetriya.

\[\begin{align*}{\left( { - x} \right)^2} + {\left( { - y} \right)^2} &= 1\\ {x^2} + {y^2} &= 1\end{align*}\]

Demak, u kelib chiqishiga nisbatan ham simmetriyaga ega.

E'tibor bering, bu aylananing markazida joylashgan va biz simmetriya haqida birinchi marta gapira boshlaganimizda ta'kidlanganidek, u uchta simmetriyaga ega.