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Agar grafik f ( x ) = ( m + 18 ) x 2 − 2 m x + 3 memotong sumbu X maka batas nilai m adalah .... ( Petunjuk : D > 0 )

Agar grafik  memotong sumbu  maka batas nilai  adalah ....
(Petunjuk: )

  1. Error converting from MathML to accessible text. 

  2. Error converting from MathML to accessible text. 

  3. size 14px minus size 14px 6 size 14px less than size 14px m size 14px less than size 14px 9 

  4. size 14px minus size 14px 9 size 14px less than size 14px m size 14px less than size 14px 6 

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R. Novianto

Master Teacher

Mahasiswa/Alumni Universitas Tanjungpura Pontianak

Jawaban terverifikasi

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Pembahasan

Agar grafik memotong sumbu maka Buat garis bilangan : uji titik (hasil negatif) batas nilai adalah Jadi, jawaban yang tepat adalah A

Agar grafik begin mathsize 14px style f left parenthesis x right parenthesis equals open parentheses m plus 18 close parentheses x squared minus 2 m x plus 3 end style memotong sumbu begin mathsize 14px style X end style maka begin mathsize 14px style D greater than 0 end style 

begin mathsize 14px style 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 D greater than 0 space space space space space space space space space space space space space space space space space space space space space b squared minus 4 a c greater than 0 open parentheses negative 2 m close parentheses squared minus 4 open parentheses m plus 18 close parentheses open parentheses 3 close parentheses greater than 0 space space space space space space space space space 4 m squared minus 12 m minus 216 greater than 0 minus negative negative negative negative negative negative negative negative negative negative space minus negative open parentheses divided by 4 close parentheses space space space space space space space space space space space space space space space m squared minus 3 m minus 54 greater than 0 space space space space space space space space space space space space space space open parentheses m minus 9 close parentheses open parentheses m plus 6 close parentheses greater than 0  text pembuat nol: end text space space space space space space space space space space space space space space space open parentheses m minus 9 close parentheses open parentheses m plus 6 close parentheses equals 0 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 m equals 9 space text atau end text space m equals negative 6  end style 

Buat garis bilangan :

uji titik begin mathsize 14px style m equals 0 rightwards arrow m squared minus 3 m minus 54 equals open parentheses 0 close parentheses squared minus 3 open parentheses 0 close parentheses minus 54 equals negative 54 end style (hasil negatif)

batas nilai size 14px m adalah begin mathsize 14px style m less than negative 6 text  atau end text space m greater than 9 end style 

Jadi, jawaban yang tepat adalah A

 

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