Roboguru

Tentukan titik potong- dan titik potong- dari fungsi-fungsi berikut: b.

Pertanyaan

Tentukan titik potong-x dan titik potong-y dari fungsi-fungsi berikut:

b. x squared plus y squared minus 8 x minus 6 y minus 11 equals 0

Pembahasan Soal:

Akan ditentukan titik potong-x, artinya y equals 0

Perhatikan persamaan berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared minus 8 x minus 6 y minus 11 end cell equals 0 row cell x squared plus left parenthesis 0 right parenthesis squared minus 8 x minus 6 left parenthesis 0 right parenthesis minus 11 end cell equals 0 row cell x squared plus 0 minus 8 x minus 0 minus 11 end cell equals 0 row cell x squared minus 8 x minus 11 end cell equals 0 end table 

maka

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 end cell equals cell fraction numerator negative b plus square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row blank equals cell fraction numerator negative left parenthesis negative 8 right parenthesis plus square root of left parenthesis negative 8 right parenthesis squared minus 4 left parenthesis 1 right parenthesis left parenthesis negative 11 right parenthesis end root over denominator 2 left parenthesis 1 right parenthesis end fraction end cell row blank equals cell fraction numerator 8 plus square root of 64 plus 44 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 8 plus square root of 108 over denominator 2 end fraction end cell row blank equals cell fraction numerator 8 plus square root of 36 cross times 3 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 8 plus 6 square root of 3 over denominator 2 end fraction end cell row blank equals cell 4 plus 3 square root of 3 end cell end table  

atau

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 2 end cell equals cell fraction numerator negative b minus square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row blank equals cell fraction numerator negative left parenthesis negative 8 right parenthesis minus square root of left parenthesis negative 8 right parenthesis squared minus 4 left parenthesis 1 right parenthesis left parenthesis negative 11 right parenthesis end root over denominator 2 left parenthesis 1 right parenthesis end fraction end cell row blank equals cell fraction numerator 8 minus square root of 64 plus 44 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 8 minus square root of 108 over denominator 2 end fraction end cell row blank equals cell fraction numerator 8 minus square root of 36 cross times 3 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 8 minus 6 square root of 3 over denominator 2 end fraction end cell row blank equals cell 4 minus 3 square root of 3 end cell end table 

Jadi titik potong-x pada x squared plus y squared minus 8 x minus 6 y minus 11 equals 0 adalah left parenthesis 4 plus 3 square root of 3 comma space 0 right parenthesis space atau space open parentheses 4 minus 3 square root of 3 comma space 0 close parentheses.

Akan ditentukan titik potong-y, artinya x equals 0

Perhatikan persamaan berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared minus 8 x minus 6 y minus 11 end cell equals 0 row cell left parenthesis 0 right parenthesis squared plus y squared minus 8 left parenthesis 0 right parenthesis minus 6 y minus 11 end cell equals 0 row cell 0 plus y squared minus 0 minus 6 y minus 11 end cell equals 0 row cell y squared minus 6 y minus 11 end cell equals 0 end table 

maka

table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 end cell equals cell fraction numerator negative b plus square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row blank equals cell fraction numerator negative left parenthesis negative 6 right parenthesis plus square root of left parenthesis negative 6 right parenthesis squared minus 4 left parenthesis 1 right parenthesis left parenthesis negative 11 right parenthesis end root over denominator 2 left parenthesis 1 right parenthesis end fraction end cell row blank equals cell fraction numerator 6 plus square root of 36 plus 44 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 6 plus square root of 80 over denominator 2 end fraction end cell row blank equals cell fraction numerator 6 plus square root of 16 cross times 5 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 6 plus 4 square root of 5 over denominator 2 end fraction end cell row blank equals cell 3 plus 2 square root of 5 end cell end table 

atau

table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 2 end cell equals cell fraction numerator negative b minus square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row blank equals cell fraction numerator negative left parenthesis negative 6 right parenthesis minus square root of left parenthesis negative 6 right parenthesis squared minus 4 left parenthesis 1 right parenthesis left parenthesis negative 11 right parenthesis end root over denominator 2 left parenthesis 1 right parenthesis end fraction end cell row blank equals cell fraction numerator 6 minus square root of 36 plus 44 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 6 minus square root of 80 over denominator 2 end fraction end cell row blank equals cell fraction numerator 6 minus square root of 16 cross times 5 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 6 minus 4 square root of 5 over denominator 2 end fraction end cell row blank equals cell 3 minus 2 square root of 5 end cell end table

Jadi titik potong-y pada x squared plus y squared minus 8 x minus 6 y minus 11 equals 0 adalah left parenthesis 0 comma space 3 plus 2 square root of 5 right parenthesis atau .left parenthesis 0 comma space 3 minus 2 square root of 5 right parenthesis

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

E. Lestari

Mahasiswa/Alumni Universitas Sebelas Maret

Terakhir diupdate 04 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Carilah solusi  bilangan real untuk setiap persamaan kuadrat berikut dan lukiska sketsa grafik penyelesaiannya!

Pembahasan Soal:

Diketahui:

x squared minus y squared equals 11 minus y squared equals 11 minus x squared space left parenthesis Kedua space ruas space dikali space left parenthesis negative right parenthesis right parenthesis y squared equals negative 11 plus x squared space... left parenthesis 1 right parenthesis 3 x squared minus 2 y squared equals 58 space... left parenthesis 2 right parenthesis

Subtitusikan persamaan (1) ke persamaan (2) yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared minus 2 open parentheses negative 11 plus x squared close parentheses end cell equals 58 row cell 3 x squared plus 22 minus 2 x squared end cell equals 58 row cell x squared end cell equals cell 58 minus 22 end cell row cell x squared end cell equals 36 row x equals cell plus-or-minus square root of 36 end cell row x equals cell plus-or-minus 6 end cell end table

Dimana, nilai x yaitu

x subscript 1 equals 6 x subscript 2 equals negative 6

Subtitusikan nilai space x subscript 1 space space dan space space x subscript 2  ke persamaan (1) yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell y squared end cell equals cell negative 11 plus x squared end cell row cell y squared end cell equals cell negative 11 plus left parenthesis 6 right parenthesis squared space end cell row cell y squared end cell equals cell negative 11 plus 36 end cell row cell y squared end cell equals 25 row y equals cell plus-or-minus square root of 25 end cell row y equals cell plus-or-minus 5 end cell row cell y subscript 1 end cell equals cell 5 space space dan space space y subscript 2 equals negative 5 end cell end table 

Jadi, solusi adalah (6,5) ,(6,-5), (-6,5) dan (-6,-5).

Grafik

  • table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus y squared end cell equals 11 row cell negative Sumbu space x comma space Ketika space y end cell equals cell 0 colon end cell row cell x squared minus left parenthesis 0 right parenthesis squared end cell equals 15 row cell x squared end cell equals 15 row x equals cell plus-or-minus square root of 15 end cell row cell x subscript 1 end cell equals cell square root of 15 space comma space space space x subscript 2 equals negative square root of 15 end cell row blank blank blank row cell negative Sumbu space y comma space ketika space x end cell equals cell 0 colon end cell row cell left parenthesis 0 right parenthesis squared minus y squared end cell equals 15 row cell negative y squared end cell equals 15 row y equals cell plus-or-minus square root of negative 15 end root end cell row blank blank cell y space tidak space punya space solusi end cell end table
  • table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared minus 2 y squared end cell equals 8 row cell negative space Sumbu space x comma space Ketika space x end cell equals cell 0 colon end cell row cell 3 left parenthesis 0 right parenthesis squared minus 2 y squared end cell equals 8 row cell negative 2 y squared end cell equals 8 row cell y squared end cell equals cell fraction numerator 8 over denominator negative 2 end fraction end cell row cell y squared end cell equals cell negative 4 end cell row y equals cell plus-or-minus square root of negative 4 end root end cell row blank blank cell straight y space tidak space punya space solusi space end cell row blank blank blank row cell negative Sumbu space x comma ketika space y end cell equals cell 0 colon end cell row cell 3 x squared minus 2 left parenthesis 0 right parenthesis squared end cell equals 8 row cell 3 x squared end cell equals 8 row cell x squared end cell equals cell 8 over 3 end cell row x equals cell plus-or-minus square root of 8 over 3 end root end cell row cell x subscript 1 end cell equals cell square root of 8 over 3 space end root space comma space x subscript 2 equals negative square root of 8 over 3 end root end cell end table

Grafik

Roboguru

Carilah semua solusi  bilangan real untuk setiap persamaan kuadrat berikut dan lukiskan sketsa grafik penyelesaianya!

Pembahasan Soal:

Diketahui:

x squared plus y squared equals 25 space... left parenthesis 1 right parenthesis x squared minus y squared equals 7 space minus y squared equals 7 minus x squared space left parenthesis Kedua space ruas space dikali space left parenthesis negative right parenthesis right parenthesis y squared equals negative 7 plus x squared space... left parenthesis 2 right parenthesis

Dengan mensubtitusikan persamaan (2) ke persamaan (1), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus y squared end cell equals 25 row cell x squared plus left parenthesis negative 7 plus x squared right parenthesis end cell equals 25 row cell 2 x squared end cell equals cell 25 plus 7 end cell row cell 2 x squared end cell equals 32 row cell x squared end cell equals cell 32 over 2 end cell row cell x squared end cell equals 16 row x equals cell plus-or-minus square root of 16 end cell row x equals cell plus-or-minus 4 end cell end table

Dimana, nilai x adalah 

x subscript 1 equals 4 space x subscript 2 equals negative 4

Mensubtitusikan x subscript 1 space space dan space space x subscript 2 ke persamaan (2) yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell y squared end cell equals cell negative 7 plus left parenthesis 4 right parenthesis squared end cell row cell space y squared end cell equals cell negative 7 plus 16 end cell row cell y squared end cell equals 9 row cell y squared end cell equals cell plus-or-minus square root of 9 end cell row y equals cell plus-or-minus 3 end cell row cell y subscript 1 end cell equals cell 3 space space space dan space space y subscript 2 equals negative 3 end cell end table

Jadi, solusi adalah (4,3, (4,-3), (-4,3) dan (-4,-3).

Grafik 

  • table attributes columnalign right center left columnspacing 0px end attributes row cell x blank squared minus y blank squared end cell equals 7 row cell space minus Sumbu space x comma space y end cell equals cell 0 colon end cell row cell space x blank squared minus 0 end cell equals cell 7 space end cell row cell x blank squared end cell equals cell 7 space end cell row x equals cell plus-or-minus square root of 7 end cell row cell space x blank subscript 1 end cell equals cell square root of 7 space comma space space x blank subscript 2 equals negative square root of 7 space end cell row cell space minus Sumbu space y comma space x end cell equals cell 0 colon space end cell row cell left parenthesis 0 to the power of right parenthesis 2 end exponent minus y squared end cell equals 7 row cell space minus y blank squared end cell equals cell 7 space end cell row y equals cell plus-or-minus square root of negative 7 end root space end cell row blank blank cell straight y space tidak space punya space solusi end cell row blank blank blank end table

Grafik

Roboguru

Carilah solusi  bilangan real untuk setiap persamaan kuadrat berikut dan lukiska sketsa grafik penyelesaiannya! -

Pembahasan Soal:

Diketahui:

2 x squared plus y squared equals 17 space... left parenthesis 1 right parenthesis x squared plus 2 y squared equals 22 x squared equals 22 minus 2 y squared space... left parenthesis 2 right parenthesis

Subtitusikan persamaan (2) ke persamaan (1), sehingga

2 left parenthesis 22 minus 2 y squared right parenthesis plus y squared equals 17 44 minus 4 y squared plus y squared equals 17 minus 3 y squared equals 17 minus 44 minus 3 y squared equals negative 27 space left parenthesis Kedua space ruas space dikali left parenthesis negative right parenthesis right parenthesis 3 y squared equals 27 y squared equals 27 over 3 y squared equals 9 y equals plus-or-minus square root of 9 y equals plus-or-minus 3 y subscript 1 equals 3 space comma space y subscript 2 equals negative 3

Nilai x subtitusikan nilai  y subscript 1 space space dan space space y subscript 2 , sehingga

Ketika comma space y equals 3 space colon x squared equals 22 minus 2 y squared x squared equals 22 minus 2 left parenthesis 3 right parenthesis squared space x squared equals 22 minus 18 x squared equals 4 x equals plus-or-minus square root of 4 x equals plus-or-minus 2 x subscript 1 equals 2 space comma space x subscript 2 equals negative 2  Ketika space y equals negative 3 colon x squared equals 22 minus 2 y squared x squared equals 22 minus 2 left parenthesis negative 3 right parenthesis squared x squared equals 22 minus 18 x squared equals 4 x equals plus-or-minus square root of 4 x equals plus-or-minus 2 x subscript 1 equals 2 space comma space x subscript 2 equals negative 2

Jadi, solusi adalah (2,3), (2,-3), (-2,3) dan (-2,-3).

Grafik

  • table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x squared plus y squared end cell equals 17 row cell negative Sumbu space x comma space y end cell equals cell 0 colon end cell row cell 2 x squared plus left parenthesis 0 right parenthesis squared end cell equals 17 row cell 2 x squared end cell equals 17 row cell x squared end cell equals cell 17 over 2 end cell row x equals cell plus-or-minus square root of 17 over 2 end root end cell row cell x subscript 1 end cell equals cell square root of 17 over 2 space comma space end root space x subscript 2 equals negative square root of 17 over 2 end root end cell row cell negative Sumbu space y comma space ketika space x end cell equals cell 0 colon end cell row cell 2 left parenthesis 0 right parenthesis squared plus y squared end cell equals 17 row cell y squared end cell equals 17 row y equals cell plus-or-minus square root of 17 end cell row cell y subscript 1 end cell equals cell square root of 17 space comma space space y subscript 2 equals negative square root of 17 end cell end table
  • table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus 2 y squared end cell equals 22 row cell negative Sumbu space x comma space y end cell equals cell 0 colon end cell row cell x squared plus 2 left parenthesis 0 right parenthesis squared end cell equals 22 row cell x squared end cell equals 22 row x equals cell plus-or-minus square root of 22 end cell row cell x subscript 1 end cell equals cell square root of 22 space comma space x subscript 2 equals negative square root of 22 end cell row blank blank blank row cell negative Sumbu space y comma space x end cell equals cell 0 colon end cell row cell 0 squared plus y squared end cell equals 22 row cell y squared end cell equals 22 row y equals cell plus-or-minus square root of 22 end cell row cell y subscript 1 end cell equals cell square root of 22 space comma space y subscript 2 equals negative square root of 22 end cell end table

Grafik

Roboguru

Carilah solusi  bilangan real untuk setiap persamaan kuadrat berikut dan lukiska sketsa grafik penyelesaiannya!

Pembahasan Soal:

Diketahui:

2 x squared plus 3 y squared equals 23 3 x squared minus 2 y squared equals 15

Misalkan : u equals x squared v equals y squared

2 u plus 3 v equals 23 space... left parenthesis 1 right parenthesis 3 u minus 2 v equals 15 space... left parenthesis 2 right parenthesis

Dengan mengeliminasikan persamaan (1) dan (2), sehingga

2 u plus 3 v equals 23 space space space left parenthesis x 3 right parenthesis space space up diagonal strike space space space 6 u end strike plus 9 v equals 69 3 u minus 2 v equals 15 space space space left parenthesis x 2 right parenthesis space space up diagonal strike space space 6 u end strike minus 4 v equals 30 space space space space space space space space space space space minus space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space top enclose space space space space space space space space space space space 13 v equals 39 space space space space space space space space space space space space space space space space space space space end enclose space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space v equals 39 over 13 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space v equals 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space

Kemudian, subtitusikan nilai v ke persamaan (1), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 u plus 3 v end cell equals 23 row cell 2 u plus 3 left parenthesis 3 right parenthesis end cell equals 23 row cell 2 u plus 9 end cell equals 23 row cell 2 u end cell equals cell 23 minus 9 end cell row cell 2 u end cell equals 14 row u equals cell 14 over 2 end cell row u equals 7 end table

Jadi,  solusi adalah (7,3)

Grafik

  • table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x squared plus 3 y squared end cell equals 23 row cell negative space Sumbu space x comma space ketika space y end cell equals cell 0 space colon end cell row cell 2 x squared plus 3 left parenthesis 0 right parenthesis squared end cell equals 23 row cell 2 x squared end cell equals 23 row cell x squared end cell equals cell 23 over 2 end cell row x equals cell plus-or-minus square root of 23 over 2 end root end cell row cell x subscript 1 end cell equals cell square root of 23 over 2 space comma space end root space x subscript 2 equals negative square root of 23 over 2 end root end cell row blank blank blank row cell negative Sumbu space y comma space ketika space x end cell equals cell 0 colon end cell row cell 2 left parenthesis 0 right parenthesis plus 3 y squared end cell equals 23 row cell 3 y squared end cell equals 23 row cell y squared end cell equals cell 23 over 2 end cell row y equals cell plus-or-minus square root of 23 over 2 end root end cell row cell y subscript 1 end cell equals cell square root of 23 over 2 space comma space end root space y subscript 2 equals negative square root of 23 over 2 end root end cell row blank blank blank end table
  •  3 x squared minus 2 y squared equals 15 minus Sumbu space x comma space ketika space y equals 0 colon 3 x squared minus 2 left parenthesis 0 right parenthesis squared equals 15 3 x squared equals 15 x squared equals 15 over 3 x squared equals 5 x equals plus-or-minus square root of 5 x subscript 1 equals square root of 5 space space space comma space x subscript 2 equals negative square root of 5 minus Sumbu space y comma space ketika space x equals 0 colon 3 left parenthesis 0 right parenthesis squared minus 2 y squared equals 15 minus 2 y squared equals 15 y squared equals negative 15 over 2 y equals plus-or-minus square root of fraction numerator negative 15 over denominator 2 end fraction end root straight y space tidak space mempunyai space penyelesaian x equals plus-or-minus square root of 5 x subscript 1 equals square root of 5 space space space comma space x subscript 2 equals negative square root of 5

Grafik

 

Roboguru

Sistem persamaan kuadrat-kuadrat dan tidak mempunyai penyelesaian. Nilai  yang memenuhi adalah ....

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 end cell equals cell y subscript 2 end cell row cell x squared minus 4 x plus m end cell equals cell 2 minus x squared end cell row cell 2 x squared minus 4 x plus m minus 2 end cell equals 0 row blank blank blank row cell Tidak space mempunyai space penyelesaian space jika space D end cell less than 0 row cell b squared minus 4 a c end cell less than 0 row cell left parenthesis negative 4 right parenthesis squared minus 4 open parentheses 2 close parentheses open parentheses m minus 2 close parentheses end cell less than 0 row cell 16 minus 8 m plus 16 end cell less than 0 row cell negative 8 m end cell less than cell negative 32 end cell row m greater than 4 row blank blank blank end table end style  

 

Jadi jawaban yang tepat adalah D

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved