Pertanyaan

Banyaknya asimtot tegak dari fungsi ada sebanyak ....

Banyaknya asimtot tegak dari fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses x squared minus 4 close parentheses open parentheses x squared minus 5 x plus 6 close parentheses over denominator open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end fraction end style$ ada sebanyak ....

H. Nufus

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

Perhatikan bahwa fungsi memiliki penyebut = 0 jika x = 2 atau x = 3 . Perhatikan bahwa Sehingga x = 2 bukan asimtot tegak dari fungsi . Selanjutnya perhatikan bahwa Sehingga x = 3 bukan asimtot tegak dari fungsi . Maka, banyaknya asimtot tegak dari fungsi ada sebanyak 0 buah.

Perhatikan bahwa fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses x squared minus 4 close parentheses open parentheses x squared minus 5 x plus 6 close parentheses over denominator open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end fraction end style$ memiliki penyebut = 0 jika x = 2 atau x = 3.

Perhatikan bahwa

$begin mathsize 14px style limit as x rightwards arrow 2 to the power of plus of invisible function application fraction numerator open parentheses x squared minus 4 close parentheses open parentheses x squared minus 5 x plus 6 close parentheses over denominator open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end fraction equals limit as x rightwards arrow 2 to the power of plus of invisible function application fraction numerator open parentheses x plus 2 close parentheses open parentheses x minus 2 close parentheses open parentheses x minus 2 close parentheses open parentheses x minus 3 close parentheses over denominator open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end fraction equals limit as x rightwards arrow 2 to the power of plus of invisible function application fraction numerator open parentheses x plus 2 close parentheses up diagonal strike open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end strike over denominator up diagonal strike open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end strike end fraction equals limit as x rightwards arrow 2 to the power of plus of invisible function application open parentheses x plus 2 close parentheses equals 2 plus 2 equals 4 end style$

Sehingga x = 2 bukan asimtot tegak dari fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses x squared minus 4 close parentheses open parentheses x squared minus 5 x plus 6 close parentheses over denominator open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end fraction end style$ .

Selanjutnya perhatikan bahwa

$begin mathsize 14px style limit as x rightwards arrow 3 to the power of plus of invisible function application fraction numerator open parentheses x squared minus 4 close parentheses open parentheses x squared minus 5 x plus 6 close parentheses over denominator open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end fraction equals limit as x rightwards arrow 3 to the power of plus of invisible function application fraction numerator open parentheses x plus 2 close parentheses open parentheses x minus 2 close parentheses open parentheses x minus 2 close parentheses open parentheses x minus 3 close parentheses over denominator open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end fraction equals limit as x rightwards arrow 3 to the power of plus of invisible function application fraction numerator open parentheses x plus 2 close parentheses up diagonal strike open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end strike over denominator up diagonal strike open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end strike end fraction equals limit as x rightwards arrow 3 to the power of plus of invisible function application open parentheses x plus 2 close parentheses equals 3 plus 2 equals 5 end style$

Sehingga x = 3 bukan asimtot tegak dari fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses x squared minus 4 close parentheses open parentheses x squared minus 5 x plus 6 close parentheses over denominator open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end fraction end style$ .

Maka, banyaknya asimtot tegak dari fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator open parentheses x squared minus 4 close parentheses open parentheses x squared minus 5 x plus 6 close parentheses over denominator open parentheses x minus 2 close parentheses squared open parentheses x minus 3 close parentheses end fraction end style$ ada sebanyak 0 buah.