Asosiy tarkibga o'tish

1.9-bo'lim: Grafik va umumiy grafiklar

Quyidagilarning har birining grafigini chizing.

  1. \(y = - \frac{2}{5}x + 3\)
    Yechim

    Bu qiyalik kesishuvi shaklidagi chiziq

    \[y = mx + b\]

    Bu holda chiziq \(\left( {0,b} \right)\) ning \(y\) kesishmasi va \(m\) qiyaligiga ega. Esda tutingki, nishab deb o'ylash mumkin

    \[m = \frac{\mbox{rise}}{\mbox{run}}\]

    Nishab salbiy bo'lsa, biz ko'tarilishni pasayish deb o'ylaymiz.

    Nishab bizga chiziqning ikkinchi nuqtasini olish imkonini beradi. Chiziq va qiyalikning istalgan nuqtasiga ega bo'lgach, biz to'g'ridan-to'g'ri harakatlanamiz yugur va yuqoriga/pastga ko'tarilish belgisiga qarab. Bu chiziqdagi ikkinchi nuqta bo'ladi.

    Bu holda biz bilamizki (0,3) chiziqdagi nuqta va qiyaligi \( - \frac{2}{5}\). Shunday qilib, (0,3) dan boshlab biz 5 ni o'ngga siljitamiz (i.e. \(0 \to 5\)) va pastga 2 (i.e. \(3 \to 1\)) chiziqdagi ikkinchi nuqta sifatida (5,1) ni olish uchun. Chiziqda ikkita nuqtani olganimizdan so'ng, biz qilishimiz kerak bo'lgan yagona narsa ikkita nuqtani chizish va ularni chiziq bilan bog'lashdir.

    Mana bu chiziqning eskizi.

    The graph of the line \(y = - \frac{2}{5}x + 3\). It is a straight line falling from left to right that crosses the y-axis at \(\left( {0,3} \right)\), passes through \(\left( {5,1} \right)\) and crosses the x-axis at \(\left( {7.5,0} \right)\).
  2. \(y = {\left( {x + 3} \right)^2} - 1\)
    Yechim

    Bu shakldagi parabola

    \[y = {\left( {x - h} \right)^2} + k\]

    Ushbu shakldagi parabolalarning uchi \(\left(h, k\right)\) da bo'ladi. Demak, parabolamizning tepasi (-3,-1) ga teng. Bundan tashqari, bu parabola ochilishini ko'rishimiz mumkin, chunki kvadratik terminning koeffitsienti ijobiy bo'ladi (va agar u salbiy bo'lsa, albatta ochiladi).

    Parabolalarning grafiklarini tuzishda \(x\)-kesishmalari mavjud bo'lsa, ular ham qulaydir. Bu holda ular mavjud bo'ladi, chunki cho'qqi \(x\) o'qi ostidadir va parabola ochiladi. Ularni topish uchun biz qilishimiz kerak bo'lgan yagona narsa.

    \[\begin{align*}{\left( {x + 3} \right)^2} - 1 & = 0\\ {\left( {x + 3} \right)^2} & = 1\\ x + 3 & = \pm 1\\ x &= - 4\,\,{\mbox{and }}\,x = - 2\end{align*}\]

    Ushbu ma'lumotlar bilan biz asosiy fikrlarni va parabolada eskizni chizishimiz mumkin. Mana eskiz.

    The graph of \(y = {\left( {x + 3} \right)^2} - 1\). It is a parabola that opens upward with vertex at \(\left( { - 3, - 1} \right)\) and x-intercepts at \(\left( { - 4,0} \right)\) and \(\left( { - 2,0} \right)\).
  3. \(y = - {x^2} + 2x + 3\)
    Yechim

    Bu ham parabolaning grafigi, faqat u umumiyroq shaklda.

    \[y = a{x^2} + bx + c\]

    Ushbu shaklda cho'qqining \(x\)-koordinatasi \(x = - \frac{b}{{2a}}\) va biz bu qiymatni tenglamaga qayta kiritish orqali \(y\)-koordinatasini olamiz. Shunday qilib, bizning parabolamiz uchun tepaning koordinatalari bo'ladi.

    \[\begin{align*} x & = - \frac{2}{{2\left( { - 1} \right)}} = 1\\ y & = - {\left( 1 \right)^2} + 2\left( 1 \right) + 3 = 4\end{align*}\]

    Demak, bu parabolaning tepasi (1,4) ga teng.

    Biz \(a\) belgisidan parabola qaysi tomonga ochilishini ham aniqlashimiz mumkin. Agar \(a\) ijobiy bo'lsa, parabola ochiladi va \(a\) manfiy bo'lsa, parabola pastga ochiladi. Bizning holatda parabola ochiladi.

    Bu shuningdek, ushbu grafikda bizda \(x\)-kesishmalar bo'lishini anglatadi. Shunday qilib, biz quyidagilarni hal qilamiz.

    \[\begin{align*} - {x^2} + 2x + 3 & = 0\\ {x^2} - 2x - 3 & = 0\\ \left( {x - 3} \right)\left( {x + 1} \right) & = 0\end{align*}\]

    Shunday qilib, biz \(x = - 1\) va \(x = 3\) da \(x\)-kesishmalariga ega bo'lamiz. E'tibor bering, hal qilish jarayonida hayotimni osonlashtirish uchun \({x^2}\) musbat koeffitsientini olish uchun hamma narsani -1 ga ko'paytirdim. Bu faktoringni osonlashtirdi.

    Mana bu parabolaning eskizi.

    The graph of \(y = - {x^2} + 2x + 3\). It is a parabola that opens downward with vertex at \(\left( {1,4} \right)\) and x-intercepts at \(\left( { - 1,0} \right)\) and \(\left( {3,0} \right)\).

    Buni oldingi masalada ishlatilgan shaklga aylantirish uchun kvadratni to'ldirishdan ham foydalanishimiz mumkin. Buni amalga oshirish uchun men birinchi navbatda -1 ni hisobga olaman, chunki \({x^2}\) koeffitsienti musbat 1 bo'lsa, kvadratni to'ldirish har doim osonroq bo'ladi.

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

    Endi \(x\) koeffitsientining yarmini oling va uni kvadratga aylantiring. Keyin buni qavs ichidagi miqdorga qo'shing va ayiring. Bu birinchi uchta shartni faktorlarga ajratilishi mumkin bo'lgan mukammal kvadratga aylantiradi.

    \[\begin{align*}y & = - \left( {{x^2} - 2x - 3} \right)\\ & = - \left( {{x^2} - 2x + 1 - 1 - 3} \right)\\ & = - \left( {{x^2} - 2x + 1 - 4} \right)\\ & = - \left( {{{\left( {x - 1} \right)}^2} - 4} \right)\end{align*}\]

    Yakuniy qadam minus belgisini ko'paytirishdir.

    \[y = - {\left( {x - 1} \right)^2} + 4\]
  4. \(f\left( x \right) = - {x^2} + 2x + 3\)
    Yechim

    Bu muammo va oldingi muammo o'rtasidagi yagona farq shundaki, bizda \(y\) o'rniga chap tomonda \(f\left( x \right)\) mavjud. Ammo shuni yodda tutingki, \(f\left( x \right)\) \(y\) yozishning shunchaki ajoyib usuli va shuning uchun \(f\left( x \right) = - {x^2} + 2x + 3\) grafigi \(y = - {x^2} + 2x + 3\) grafigiga o'xshaydi.

    Shunday qilib, ish uchun oldingi muammoni ko'ring. To'liqlik uchun bu erda grafik.

    The graph of \(y = - {x^2} + 2x + 3\). It is a parabola that opens downward with vertex at \(\left( {1,4} \right)\) and x-intercepts at \(\left( { - 1,0} \right)\) and \(\left( {3,0} \right)\).
  5. \(x = {y^2} - 6y + 5\)
    Yechim

    Aksariyat odamlar \(y = f(x)\) ko'rinishidagi funktsiyalar bilan ishlashga qodir Algebra kursidan chiqadilar. Biroq, Calculus kursida ishlashingiz kerak bo'lgan ko'plab funktsiyalar \(x = f(y)\) shaklida bo'lib, ular bilan faqat shu shaklda ishlash oson. Demak, bu formadagi funksiyalar bilan ishlashga odatlanish kerak.

    Ushbu turdagi funksiyalarning yoqimli tomoni shundaki, agar siz \(y = f(x)\) ko'rinishidagi funktsiyalar bilan shug'ullanishingiz mumkin bo'lsa, unda siz \(x = f(y)\) ko'rinishidagi funktsiyalar bilan shug'ullanishingiz mumkin.

    So,

    \[y = {x^2} - 6x + 5\]

    ochiladigan parabola bo'lib, uchi (3,-4) ga teng.

    Xo'sh, bizning funktsiyamiz shaklda

    \[x = a{y^2} + by + c\]

    va bu \(a\) belgisiga qarab chapga yoki o'ngga ochiladigan parabola (\(a\) musbat bo'lsa o'ngga, \(a\) manfiy bo'lsa chapga). Tenglikning \(y\)-koordinatasi \(y = - \frac{b}{{2a}}\) tomonidan berilgan va biz \(x\)-koordinatasini tenglamaga ulab topamiz.

    Bizning funktsiyamiz o'ngga ochiladigan (\(a\) musbat) va cho'qqisi (-4,3) bo'lgan paraboladir. Buni grafik qilish uchun bizga \(y\)-kesishmalar kerak bo'ladi. Biz ularni xuddi oldingi ikkita muammoda \(x\)-kesishmalarini topgandek topamiz.

    \[\begin{align*}{y^2} - 6y + 5 & = 0\\ \left( {y - 5} \right)\left( {y - 1} \right) & = 0\end{align*}\]

    Parabolada \(y = 1\) va \(y = 5\) da \(y\)-kesishmalar bo'ladi. Mana grafikning eskizi.

    The graph of \(x = {y^2} - 6y + 5\). It is a parabola that opens to the right with vertex at \(\left( { - 4,3} \right)\) and y-intercepts at \(\left( {0,1} \right)\) and \(\left( {0,5} \right)\).
  6. \({x^2} + {\left( {y + 5} \right)^2} = 4\)
    Yechim

    Bu uning standart shaklidagi doiradir

    \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]

    Doiralar bu shaklda bo'lsa, biz markazni, \(\left( {h,k} \right)\) va radiusni, \(r\) ni osongina aniqlashimiz mumkin. Ularga ega bo'lganimizdan so'ng, biz aylana grafigini oddiygina markazdan boshlab va \(r\) yordamida o'ngga, chapga, yuqoriga va pastga siljitish orqali mos ravishda eng o'ng, eng chap, eng yuqori va pastki eng ko'p nuqtalarni olishimiz mumkin.

    Bizning doiramizning markazi (0, -5) va radiusi 2 ga teng. Mana bu doiraning eskizi.

    The graph of \({x^2} + {\left( {y + 5} \right)^2} = 4\). It is a circle centered at \(\left( {0, - 5} \right)\) with a radius of 2, so it runs from \(\left( { - 2, - 5} \right)\) to \(\left( {2, - 5} \right)\) across and from \(\left( {0, - 7} \right)\) to \(\left( {0, - 3} \right)\) top to bottom.
  7. \({x^2} + 2x + {y^2} - 8y + 8 = 0\)
    Yechim

    Bu ham aylana, lekin u standart shaklda emas. Buni grafik qilish uchun biz uni standart shaklga keltirishimiz kerak. Biroq, buni qilish juda oson. Bizga faqat \(x\) va \(y\) da kvadratni to‘ldirish kifoya.

    \[\begin{align*}{x^2} + 2x + {y^2} - 8y + 8 & = 0\\ {x^2} + 2x + 1 - 1 + {y^2} - 8y + 16 - 16 + 8 & = 0\\ {\left( {x + 1} \right)^2} + {\left( {y - 4} \right)^2} & = 9\end{align*}\]

    Shunday qilib, markaz (-1, 4) va radius 3 ga o'xshaydi. Mana aylananing eskizi.

    The graph of \({x^2} + 2x + {y^2} - 8y + 8 = 0\). It is a circle centered at \(\left( { - 1,4} \right)\) with a radius of 3, so it runs from \(\left( { - 4,4} \right)\) to \(\left( {2,4} \right)\) across and from \(\left( { - 1,1} \right)\) to \(\left( { - 1,7} \right)\) top to bottom.
  8. \(\displaystyle \frac{{{{\left( {x - 2} \right)}^2}}}{9} + 4{\left( {y + 2} \right)^2} = 1\)
    Yechim

    Bu ellips. Ellipsning standart shakli

    \[\frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\]

    Bu markazi \(\left(h, k\right)\) bo'lgan ellips bo'lib, o'ngdagi eng va chapdagi eng ko'p nuqta markazdan \(a\) masofada, eng yuqori va pastki qismi esa markazdan \(b\) masofada joylashgan.

    Bu muammo uchun ellips markazi (2, -2) va \(a = 3\) va \(b = \frac{1}{2}\) mavjud. Mana ellipsning eskizi.

    The graph of \(\frac{{{{\left( {x - 2} \right)}^2}}}{9} + 4{\left( {y + 2} \right)^2} = 1\). It is an ellipse centered at \(\left( {2, - 2} \right)\) that is long and flat, reaching out 3 units to the left and right of the center to \(\left( { - 1, - 2} \right)\) and \(\left( {5, - 2} \right)\) but only \(\frac{1}{2}\) a unit above and below the center.
  9. \(\displaystyle \frac{{{{\left( {x + 1} \right)}^2}}}{9} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1\)
    Yechim

    Bu giperbola. Giperbolaning ikkita standart shakli mavjud.

    Shakl \(\displaystyle \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1\) \(\displaystyle \frac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} - \frac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} = 1\)
    Markaz \(\left(h, k\right)\) \(\left(h, k\right)\)
    Ochiladi O'ngga va chapga ochiladi Yuqoriga va pastga ochiladi
    Cho'qqilar \(a\) birliklari o'ng va chap
    markazdan    		 \(b\) birliklari yuqoriga va pastga
    markazdan
    Asimptotlarning qiyaligi \(\displaystyle \pm \frac{b}{a}\) \(\displaystyle \pm \frac{b}{a}\)

    Xo'sh, bularning barchasi nimani anglatadi? Birinchidan, atamalardan biri ijobiy, ikkinchisi esa salbiy ekanligiga e'tibor bering. Bu giperbolaning ikki qismi qaysi tomonga ochilishini aniqlaydi. Agar \(x\) atamasi ijobiy bo'lsa, giperbola chapga va o'ngga ochiladi. Xuddi shunday, agar \(y\) atamasi ijobiy bo'lsa, giperbola yuqoriga va pastga ochiladi.

    Ikkalasi ham bir xil "markazga" ega. Giperbolalarda aylana va ellipslarning markazlari bor degan ma'noda markaz yo'q. Markaz giperbolaning grafigini chizishning boshlang'ich nuqtasidir. U cho'qqilarga qanday erishish va asimptotalarni qanday o'rnatish kerakligini aytadi.

    Giperbolaning asimptotalari markazda kesishgan va yuqorida sanab o'tilgan qiyaliklarga ega bo'lgan ikkita chiziqdir. Markazdan uzoqroqqa o'tsangiz, grafik ularning asimptotalariga yaqinlashadi.

    Bu erda keltirilgan tenglama uchun giperbola chapga va o'ngga ochiladi. Uning markazi (-1, 2). Ikki uchlari (-4, 2) va (2, 2). Asimptotalar \( \pm \frac{2}{3}\) qiyaliklarga ega bo'ladi.

    Mana bu giperbolaning eskizi.

    The graph of \(\frac{{{{\left( {x + 1} \right)}^2}}}{9} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1\). It is a hyperbola centered at \(\left( { - 1,2} \right)\) that opens left and right with vertices at \(\left( { - 4,2} \right)\) and \(\left( {2,2} \right)\). The two dashed asymptotes cross at the center and have slopes of \( \pm \frac{2}{3}\) and the two branches of the hyperbola get closer and closer to them as you move away from the center.
  10. \(y = \sqrt x \)
    Yechim

    Bu muammo uchun juda ko'p narsa yo'q. Bu shunchaki kvadrat ildiz va bu grafik bo'lib, siz vaqti-vaqti bilan eskiz qilishingiz kerak bo'ladi, shuning uchun bu erda. Shuni yodda tutingki, siz hech qanday salbiy \(x\) ni kvadrat ildizga ulay olmaysiz. Mana grafik.

    The graph of \(y = \sqrt x \) on \(0 \le x \le 16\). It starts at the origin and increases, rising steeply at first and then flattening out as it goes right, reaching \(\left( {16,4} \right)\) at the right end.
  11. \(y = {x^3}\)
    Yechim

    Hech qanday munozaraga muhtoj bo'lmagan yana bir oddiy grafik, bu shunchaki uning eskizini chizishingiz mumkinligiga ishonch hosil qilish uchun.

    The graph of \(y = {x^3}\) on \( - 2 \le x \le 2\). It comes up from the lower left, flattens out as it passes through the origin and then increases steeply to the upper right.
  12. \(y = \left| x \right|\)
    Yechim

    Agar siz qanday baholashni eslasangiz, bu oxirgi eskiz unchalik qiyin emas Mutlaq qiymat funktsiyalari. Mana buning eskizi.

    The graph of \(y = \left| x \right|\). It is a “V” shaped graph with the point of the “V” at the origin and the two halves forming 45 degree angles with the positive and negative x-axis.