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Jika y=x2+2x−24 dan x=4 maka y=....

Pertanyaan

Jika y=x2+2x24 dan x=4 maka y=....

  1. 0

  2. 1

  3. 2

  4. 3

  5. 4

Pembahasan Soal:

Untuk menjawab pertanyaan di atas, substitusikan nilai x=4 ke y=x2+2x24, diperoleh hasil

y=====x2+2x2442+242416+82424240

Jadi, jika y=x2+2x24 dan x=4 maka y=0.

Oleh karena itu, pilihan jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Nikmah

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 12 September 2021

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Pertanyaan yang serupa

Untuk   dan   bilangan real yang memenuhi  , carilah batasan nilai dari setiap variabel berikut. b.

Pembahasan Soal:

Perhatikan kembali

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 2 x y plus 2 y squared end cell equals 2 row cell x squared minus 2 x y plus y squared end cell equals cell 2 minus y squared end cell row cell left parenthesis x minus y right parenthesis squared end cell equals cell 2 minus y squared end cell row cell x minus y end cell equals cell plus-or-minus square root of 2 minus y squared end root end cell row x equals cell y plus-or-minus square root of 2 minus y squared end root space space... left parenthesis 1 right parenthesis end cell end table

Batasan nilai y.

2 minus y squared greater or equal than 0 minus y squared greater or equal than 0 minus 2 minus y squared greater or equal than negative 2 space left parenthesis Kedua space ruas space dikali space left parenthesis negative right parenthesis right parenthesis y squared greater or equal than 2 y greater or equal than plus-or-minus square root of 2  y subscript 1 equals square root of 2 y subscript 2 equals negative square root of 2

Jadi, batasan nilai y  yaitu antara negative square root of 2 space space space dan space space space square root of 2.

0

Roboguru

memenuhi sistem persamaan  dan . Nilai  adalah ...

Pembahasan Soal:

Diasumsikan  begin mathsize 14px style y greater or equal than negative x squared minus 4 x plus 5 end style dan begin mathsize 14px style y less or equal than negative 2 x plus 6 end style merupakan sistem persamaan.

Diketahui:

begin mathsize 14px style y equals negative x squared minus 4 x plus 5 end style 
begin mathsize 14px style y equals negative 2 x plus 6 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals y row cell negative x squared minus 4 x plus 5 end cell equals cell negative 2 x plus 6 end cell row cell negative x squared plus 2 x minus 4 x plus 5 minus 6 end cell equals 0 row cell negative x squared minus 2 x minus 1 end cell equals 0 row cell x squared plus 2 x plus 1 end cell equals 0 row cell left parenthesis x plus 1 right parenthesis squared end cell equals 0 row cell space x subscript 1 end cell equals cell negative 1 end cell end table end style 

Substitusikan nilai begin mathsize 14px style x end style ke begin mathsize 14px style y equals negative 2 x plus 6 end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 end cell equals cell negative 2 x plus 6 end cell row blank equals cell negative 2 left parenthesis negative 1 right parenthesis plus 6 end cell row blank equals cell 2 plus 6 end cell row blank equals 8 end table end style

Nilai begin mathsize 14px style x subscript 1 plus y subscript 1 end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus y subscript 1 end cell equals cell negative 1 plus 8 end cell row blank equals 7 end table end style

Jadi, nilai begin mathsize 14px style x subscript 1 plus y subscript 1 end style dari sistem persamaan sistem persamaan begin mathsize 14px style y equals negative x squared minus 4 x plus 5 end style dan begin mathsize 14px style y equals negative 2 x plus 6 end style adalah begin mathsize 14px style 7 end style.

0

Roboguru

Untuk   dan   bilangan real yang memenuhi  , carilah batasan nilai dari setiap variabel berikut. a.

Pembahasan Soal:

Perhatikan kembali

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 2 x y plus 2 y squared end cell equals 2 row cell x squared minus 2 x y plus y squared end cell equals cell 2 minus y squared end cell row cell left parenthesis x minus y right parenthesis squared end cell equals cell 2 minus y squared end cell row cell x minus y end cell equals cell plus-or-minus square root of 2 minus y squared end root end cell row x equals cell y plus-or-minus square root of 2 minus y squared end root space space... left parenthesis 1 right parenthesis end cell end table

Batasan nilai x.

1. x  minimum , ketika  y equals negative 1.

Selanjutnya, subtitusikan nilai y  ke persamaan (1), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell negative 1 plus-or-minus square root of 2 minus left parenthesis negative 1 right parenthesis squared end root end cell row x equals cell negative 1 plus-or-minus square root of 2 minus 1 end root end cell row x equals cell negative 1 plus-or-minus square root of 1 end cell row x equals cell negative 1 plus-or-minus 1 end cell row blank blank blank row cell x subscript 1 end cell equals cell negative 1 minus 1 end cell row blank equals cell negative 2 end cell row cell x subscript 2 end cell equals cell negative 1 plus 1 end cell row blank equals 0 end table

2.  x  maximum , ketika  y equals 1.

Selanjutnya, subtitusikan nilai y  ke persamaan (1), sehingga

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell 11 plus-or-minus square root of 2 minus left parenthesis 1 right parenthesis squared end root end cell row x equals cell 1 plus-or-minus square root of 2 minus 1 end root end cell row x equals cell 1 plus-or-minus square root of 1 end cell row x equals cell 1 plus-or-minus 1 end cell row blank blank blank row cell x subscript 1 end cell equals cell 1 plus 1 end cell row blank equals 2 row cell x subscript 2 end cell equals cell 1 minus 1 end cell row blank equals 0 end table

Jadi, batasan nilai x  yaitu antara  -2 dan 2.

0

Roboguru

Titik  terletak pada grafik . Jika  maka nilai  adalah ....

Pembahasan Soal:

Karena titik begin mathsize 14px style open parentheses p comma q close parentheses end style terletak pada grafik begin mathsize 14px style y equals q x squared minus open parentheses q squared minus 1 close parentheses x plus 90 end style maka titik  begin mathsize 14px style open parentheses p comma q close parentheses end style disubstitusikam pada grafik begin mathsize 14px style y equals q x squared minus open parentheses q squared minus 1 close parentheses x plus 90 end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row q equals cell q p squared minus open parentheses q squared minus 1 close parentheses p plus 90 end cell row blank left right double arrow cell q equals q p squared minus q squared p plus p plus 90 end cell end table end style 

Diketahui pula begin mathsize 14px style q minus p equals 9 end style maka begin mathsize 14px style q equals 9 plus p end style diperoleh

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row q equals cell q p squared minus q squared p plus p plus 90 end cell row blank left right double arrow cell open parentheses p plus 9 close parentheses equals open parentheses p plus 9 close parentheses p squared minus open parentheses p plus 9 close parentheses squared p plus p plus 90 end cell row blank left right double arrow cell 9 minus 90 equals p cubed plus 9 p squared minus open parentheses p squared plus 18 p plus 81 close parentheses p plus p minus p end cell row blank left right double arrow cell negative 81 equals p cubed plus 9 p squared minus p cubed minus 18 p squared minus 81 p end cell row blank left right double arrow cell 0 equals negative 9 p squared minus 81 p plus 81 end cell row blank left right double arrow cell 0 equals p squared plus 9 p minus 9 end cell end table end style 

Diperoleh 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell p subscript 1 comma 2 end subscript end cell equals cell fraction numerator negative 9 plus-or-minus square root of 9 squared minus 4 times 1 times open parentheses negative 9 close parentheses end root over denominator 2 end fraction end cell row blank equals cell fraction numerator negative 9 plus-or-minus square root of 81 plus 36 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator negative 9 plus-or-minus square root of 117 over denominator 2 end fraction space end cell row blank equals cell fraction numerator negative 9 plus-or-minus 3 square root of 13 over denominator 2 end fraction space end cell row blank equals cell negative 9 over 2 plus-or-minus 3 over 2 square root of 13 end cell end table end style

Untuk 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row p equals cell negative 9 over 2 plus 3 over 2 square root of 13 end cell row blank rightwards double arrow cell q equals 9 over 2 plus 3 over 2 square root of 13 end cell end table end style 

Untuk 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row p equals cell negative 9 over 2 minus 3 over 2 square root of 13 end cell row blank rightwards double arrow cell q equals 9 minus 9 over 2 minus 3 over 2 square root of 13 end cell row blank equals cell 9 over 2 minus 3 over 2 square root of 13 end cell end table end style

Dengan demikian diperoleh 

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell 1 over p minus 1 over q end cell equals cell fraction numerator q minus p over denominator p q end fraction end cell row blank equals cell fraction numerator open parentheses 9 over 2 plus 3 over 2 square root of 13 close parentheses minus open parentheses negative begin display style 9 over 2 end style plus begin display style 3 over 2 end style square root of 13 close parentheses over denominator open parentheses negative 9 over 2 plus 3 over 2 square root of 13 close parentheses open parentheses 9 over 2 plus 3 over 2 square root of 13 close parentheses end fraction end cell row blank equals cell fraction numerator 9 over 2 plus 3 over 2 square root of 13 plus 9 over 2 minus 3 over 2 square root of 13 over denominator negative begin display style 81 over 4 end style minus begin display style 27 over 4 end style square root of 13 plus begin display style 27 over 4 end style square root of 13 plus begin display style fraction numerator 9 times 13 over denominator 4 end fraction end style end fraction end cell row blank equals cell fraction numerator 9 over denominator begin display style fraction numerator negative 81 plus 117 over denominator 4 end fraction end style end fraction end cell row blank equals cell 9 times 4 over 36 end cell row blank equals 1 end table end style

Dengan demikian diperoleh nilai  begin mathsize 14px style 1 over p minus 1 over q end style adalah 1.

Jadi jawaban yang tepat adalah D.

0

Roboguru

Jika grafik fungsi menyinggung sumbu X, nilai m yang memenuhi adalah ....

Pembahasan Soal:

begin mathsize 14px style bullet Grafik space yang space menyinggung space sumbu space straight X space akan space memiliki space straight D equals 0 comma space dengan space straight D equals straight b squared minus 4 ac  Grafik space fungsi space y equals 3 x squared plus left parenthesis m minus 2 right parenthesis x plus 3 space dengan space straight a equals 3 comma space straight b equals straight m minus 2 comma space straight c equals 3 comma space  menyinggung space sumbu space straight X    straight D equals 0  straight b squared minus 4 ac equals 0  open parentheses straight m minus 2 close parentheses squared minus 4 open parentheses 3 close parentheses open parentheses 3 close parentheses equals 0  straight m squared minus 4 straight m plus 4 minus 36 equals 0  straight m squared minus 4 straight m minus 32 equals 0  open parentheses straight m plus 4 close parentheses open parentheses straight m minus 8 close parentheses equals 0  straight m equals negative 4 space atau space straight m equals 8 end style

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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