Pertanyaan

Asimtot datar dari fungsi adalah ....

Asimtot datar dari fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses 2 x minus 1 close parentheses squared over denominator 3 x squared plus 7 x minus 5 end fraction end style$ adalah ....

H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

Perhatikan bahwa Karena maka asimtot datar dari fungsi adalah

Perhatikan bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow straight infinity of invisible function application fraction numerator open parentheses 2 x minus 1 close parentheses squared over denominator 3 x squared plus 7 x minus 5 end fraction end cell equals cell limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator open parentheses 2 x minus 1 close parentheses squared over denominator 3 x squared plus 7 x minus 5 end fraction times fraction numerator 1 over x squared over denominator 1 over x squared end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator open parentheses 2 x minus 1 close parentheses squared over denominator 3 x squared plus 7 x minus 5 end fraction times fraction numerator 1 over x squared over denominator 1 over x squared end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow straight infinity of invisible function application open parentheses fraction numerator open parentheses 2 x minus 1 close parentheses squared over denominator 3 x squared plus 7 x minus 5 end fraction times fraction numerator open parentheses 1 over x close parentheses squared over denominator 1 over x squared end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow straight infinity of invisible function application fraction numerator open parentheses open parentheses 2 x minus 1 close parentheses times 1 over x close parentheses squared over denominator 3 plus 7 over x minus 5 over x squared end fraction end cell row blank equals cell limit as x rightwards arrow straight infinity of invisible function application fraction numerator open parentheses 2 minus 1 over x close parentheses squared over denominator 3 plus 7 over x minus 5 over x squared end fraction end cell row blank equals cell fraction numerator open parentheses 2 minus 0 close parentheses squared over denominator 3 plus 0 minus 0 end fraction end cell row blank equals cell 2 squared over 3 end cell row blank equals cell 4 over 3 end cell end table end style$

Karena $begin mathsize 14px style limit as x rightwards arrow straight infinity of invisible function application fraction numerator open parentheses 2 x minus 1 close parentheses squared over denominator 3 x squared plus 7 x minus 5 end fraction equals 4 over 3 end style$ maka asimtot datar dari fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses 2 x minus 1 close parentheses squared over denominator 3 x squared plus 7 x minus 5 end fraction end style$ adalah