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10. Jika garis singgung kurva y = 9 – x2 di titik P(a, b) dengan b > 0 memotong sumbu x di
titik Q(-5, 0), maka ab adalah...
(A) -10 (B) -8 (C) 0 (D) 8 |
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(E) 10
Jawaban: B P(a,b) paday=9-x2 b = 9 - a2 gradien m = y' ® m = -2x di titik (a, b) |
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m = -2a
P.G.S® y-b=-2a(x-a)
y - (9 - a2) = -2a(x - a) melalui Q(-5, 0) ® -9 + a2 = -2a(-5 - a) -9 + a2 = 10a + 2a2 0 = a2 + 10a + 9 |
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0 = (a + 1)(a + 9)
a=-1 Ú a=-9
b = 8 b = -55
b>0 (TM)
jadi a · b = -8 |
Soal SBMPTN 2018 + Pembahasan No. 10
Reviewed by Satuan Kejuruan
on
April 20, 2019
Rating:
Reviewed by Satuan Kejuruan
on
April 20, 2019
Rating:
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