Question

Tolong bantuannya ya!!!!!!!!!!!!!!!!!!!!1

Answer 1

Tips perngerjaan.
Nomor 3.
y = x² - 3
Refleksi dengan sumbu x menyebabkan:
x ⇒ x
y ⇒ -y
Menyebabkan:
-y = x² - 3
y = -x² + 3
Dilatasi dengan titik pusat dengan skala 2.
½y = -(½x)² + 3
½y = -¼x² + 3
y = -½ x² + 6
y = 6 - ½ x² [D]

Nomor 4.
Ambil inversnya:
Dari x = 2 + y - y², dilatasi ½
2x = 2 + 2y - (2y)²
2x = 2 + 2y - 4y²
x = 1 + y - 2y²
x = -2y² + y + 1
Invers rotasi ½π adalah -½π
Menyebabkan:
x' ⇒ y, sehingga x = -y'
y' ⇒ -x, sehingga y = x'
Didapat:
-y = -2x² + x + 1
y = 2x² - x - 1 [D]

Nomor 5.
y = x - 3
Ambil titik sembarang di garis ini, misal:
(0,-3) dan (3,0)
Dilatasi (2,1) skala 4
Titik (0,-3) bayangan:
= 4(0-2,-3-1)+(2,1)
= (-8,-16) + (2,1) = (-6,-15)
Titik (3,0) bayangan:
= 4(3-2,0-1) + (2,1)
= (4,-4) + (2,1) = (6,-3)

Hubungkan kedua titik pada satu persamaan garis:
[y+15]/[-3+15] = [x+6]/[6+6]
[y + 15] / 12 = [x + 6]/12
y + 15 = x + 6
x - y + 6 - 15 = 0
x - y - 9 = 0 [N/A]

Answer 2

Y = x² - 3

misal x = 0
y = -3

misal y = 0
x² - 3 = 0
x² = 3
x = +√3
x = -√3

dicerminkan :

x' = x
y' = -y

titik pertama :

x' = 0
y' = 3

(x',y') = (0,3)

titik kedua :

x' = √3
y' = 0

(x',y') = (√3,0)

titik ketiga :

x' = -√3
y' = 0

(x',y') = (-√3,0)

didilatasi

titik pertama :

x' = 2(0)
y' = 2(3)

(x',y') = (0,6)

titik kedua :

x' = 2(√3)
y' = 2(0)

(x',y') = (2√3,0)

titik ketiga :

x' = 2(-√3)
y' = 2(0)

(x',y') = (-2√3,0)

persamaan kurva = ax² + bx + c

(0, 6)
a(0)² + b(0) + c = 6
c = 6

(2√3,0)
a(2√3)² + b(2√3) + c = 0
12a + 2√3b + c = 0
12a + 2√3b + 6 = 0
12a + 2√3b = -6

(-2√3,0)
a(-2√3)² + b(-2√3) + c = 0
12a - 2√3b + c = 0
12a - 2√3b + 6 = 0
12a - 2√3b = -6

12a + 2√3b = -6
12a - 2√3b = -6_+
24a = -12
a = -12/24
a = -1/2

12a + 2√3b = -6
12(-1/2) + 2√3b = -6
-6 + 2√3b = -6
2√3b = 6 - 6
2√3b = 0
b = 0

ax² + bx + c = -1/2x² + 6
y = -1/2x² + 6

==========

x = 2 + y - y²
misal y = 0
x = 2

misal x = 0

-y² + y + 2 = 0
y² - y - 2 = 0
(y +1)(y - 2)

y = -1 dan y = 2

(2, 0) (0, -1) dan (0, 2)

dilatasi matriks :

(x') = (k 0)(x)
(y') = (0 k)(y)

(0) = (2 0)(x)
(2) = (0 2)(y)

0 = 2x
2 = 2y
x = 0
y = 1
= (0, 1)

(0) = (2 0)(x)
(-1) = (0 2)(y)

0 = 2x
-1 = 2y

x = 0
y = -1/2
= (0, -1/2)

(2) = (2 0)(x)
(0) = (0 2)(y)

2 = 2x
0 = 2y

x = 1, y = 0

= (1, 0)

(0, 1) (0, -1/2) (1,0)

rotasi sudut 1/2 = 180 derajat

x' = x cosα - y sinα
y' = x sinα + y cosα

0 = x cos180 - y sin180
1 = x sin180 + y cos180

0 = -x
1 = -y

x = 0
y = -1
= (0, -1)

0 = x cos180 - y sin180
-1/2 = x sin180 + y cos180

0 = -x
-1/2 = -y

x = 0, y = 1/2
= (0, 1/2)

0 = x cos180 - y sin180
1 = x sin180 + y cos180

1 = -x
0 = -y

x = -1
y = 0

= (-1, 0)

= (0, -1) (0,-1/2) (-1,0)

ay² + by + c = x
c = -1

a - b = 1
1/4a - 1/2b = 1

a - 2b = 4

a - b = 1
a - 2b = 4

b = -3

a - b = 1
a + 3 = 1
a = -2

penyelesaian sama seperti nomor di atas. ax² + bx + c =

nomor 5 juga penyelesaian sama seperti nomor 3, tanpa refleksi.