Kunci Jawaban Matematika Kelas 9 Halaman 102 dan 103 Latihan 2.3 Sumbu Simetri dan Titik Optimum

maka:
a. y = 2x^2 – 5x
a = 2 b = -5 c = 0

Jadi sumbu simetri:
x = -b/2a
= -(-5)/2(2)
= 5/4

b. y = 3x^2 + 12x
a = 3 b = 12 c = 0

Jadi sumbu simetri:
x = -b/2a
= -(12)/2(3)
= -12/6
= -2

c. y = -8x^2 – 16x – 1
a = -8 b = -16 c = -1

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

Baca juga: Kunci Jawaban Matematika Kelas 10 Halaman 31 dan 32 Kurikulum Merdeka Eksponen dan Logaritma

Soal nomor 2

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

Maka:
-D/4a = -(b^2 – 4ac) / 4c
-(242 – 4 (-6) (-19) / 4(-6)
= -(576 – 456)/-24 -(120)/-24
= 5

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

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

c. y = -3/4x^2 + 7x – 18
a = -3/4 b = 7 c = -18

Maka:
-D/4a = -(b^2 – 4ac) / 4c
-(7^2 – 4(-3/4) (-18)) / 4 (-3/4)
=-(49-54) / -3
5/-3

Soal nomor 3

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