Kunci Jawaban Matematika Kelas 9 Halaman 102 Cara Hitung Sumbu Simetri dan Titik Optimum

Jadi sumbu simetri:
x = -b/2a
= -(-16) / 2(-8)
= 16/-16
= -1

Soal nomor 2

a. y = -6x^2 + 24x – 19
a = -6 b = 24 c = -19

Maka:
-D/4a = -(b² – 4ac) / 4c
-(24² – 4 (-6) (-19) / 4(-6)
= -(576 – 456)/-24 -(120)/-24
= 5

b. y = 2/5x² – 3x + 15
a = 2/5 b = -3 c = 15

Maka:
-D/4a = -(b² – 4ac) / 4c
-(-32 – 4(2/5) 15) / 4. 2/5
-(9-24)/8/5
15/ 8/5
= 15.5/8
= 75/8

c. y = -3/4x² + 7x – 18
a = -3/4 b = 7 c = -18

Maka:
-D/4a = -(b² – 4ac) / 4c
-(7² – 4(-3/4) (-18)) / 4 (-3/4)
=-(49-54) / -3
5/-3

Soal nomor 3

a. y = 2x² + 9x
Sumbu x saat y
2x² + 9x = 0
x (2x + 9) = 0
maka:
x = 0 atau 2x + 9 = 0
2x = -9
x = -9/2

jadi titik (0,0) ; (-9/2,0)
sumbu y saat x = 0
y = 2x² + 9x
y = 2(0)² + 9(0)
y = 0
Maka titik (0,0)

Jadi titik baliknya adalah
xa = -b/2a = -9/2(2) = -9/4
ya = -b² – 4ac / 4a
ya = -b² – 4ac / 4a
ya = - ( 9² – 4.2.0) / 4(2)
ya = - (81 – 0) / 8
ya = -81 / 8
Koordinat titik balik:
(-9/4, -81/8)
(-2,25 ; -10,125)

b. y = 8x² – 16x + 6
Sumbu x ketika y = 0
8x^2 – 16x + 6 = 0
(4x – 2)(2x – 3) = 0

Maka:
4x – 2 = 0
4x = 2
x = 2/4 = 1/2

dan 2x – 3 = 0
2x = - 3
x = -3/2
Maka titik (1/2,0) ; (-3/2,0)