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SMK Teknologi dan Rekayasa

Soal 21 Grafik Fungsi Kuadrat


Koordinat\ titik\ puncak\ dari\ grafik\ fungsi\ \ y=2x^{2}-4x-12\\adalah\ \ ....\\a.\ \ (-1,\ -12)\ \ \ \ b.\ \ (2,\ -12)\ \ \ \ c.\ \ (-1,\ -14)\ \ \ \ d.\ \ (2,\ -12)\ \ \ \ e.\ \ (1,\ -14)
Jawaban: e

Grafik\ fungsi\ \ y=2x^{2}-4x-12\ \ berupa\ parabola.
Dari\ persamaan\ fungsi\ diperoleh:\ a=2,\ b=-4,\ c=-12
Absis\ titik\ puncak:
\begin{array}{rcl}x&=&-\frac{b}{2a}\\\\&=&-\frac{-4}{2.2}\\\\&=&-\frac{-4}{4}\\\\&=&-(-1)\\\\&=&1\end{array}
Ordinat\ titik\ puncak:
\begin{array}{rcl}y&=&2(1)^{2}-4(1)-12\\&=&2.1-4-12=2-16\\&=&-14\end{array}
Koordinat\ titik\ puncak\ adalah\ (1,\ -14).

About Kalakay

Guru Matematika SMK

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One thought on “Soal 21 Grafik Fungsi Kuadrat

  1. Pusing Pak gak ngerti, udah lama gak belajar!!!!
    Tapi bagus juga lho blognya

    Posted by Ahmad | 6 Maret 2010, 11:03 PM

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