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Fungsi Kuadrat

Authored by Rudolf Luhukay

Mathematics

University

Used 7+ times

Fungsi Kuadrat
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10 questions

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1.

OPEN ENDED QUESTION

3 mins • Ungraded

1. f(x) = 4x² + 3x + 8. Hitunglah nilai a + 2b + 3c!

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Answer explanation

Diketahui nilai a = 4, b = 3, c = 8

= a + 2b + 3c

= 4 + 2(3) + 3(8)

= 4 + 6 + 24

= 34

2.

OPEN ENDED QUESTION

3 mins • Ungraded

f(x) = 3x² - 2x + 5 memiliki bentuk sesuai dengan bentuk f(x) = ax² + bx + c. Hitunglah nilai 2a + 3b + 4c!

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Answer explanation

Diketahui nilai a = 3, b = -2, c = 5

= 2a + 3b + 4c

= 2(3) + 3(-2) + (4 x 5)

= 6 - 6 + 20

= 20

3.

OPEN ENDED QUESTION

3 mins • Ungraded

3. Diketahui fungsi f(x) = x² + 4x + 5. Hitunglah bayangangan untuk nilai x = 3

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Answer explanation

f(x) = x² + 4x + 5

= f(3) = 3² + 4(3) + 5

= f(3) = 9 + 12 + 5

= f(3) = 26

4.

OPEN ENDED QUESTION

3 mins • Ungraded

Diketahui fungsi kuadrat y = 2x2 + 4x - 6. Tentukan sumbu simetrinya!

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Answer explanation

x = -(b/2a)

= x = -(4/2x2)

= x = -(4/4) = -1

5.

OPEN ENDED QUESTION

3 mins • Ungraded

Diketahui fungsi kuadrat y = 3x2 + 6x + 5. Tentukan titik puncaknya!

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Answer explanation

x = -(b/2a)

x = -(6/2x3)

x = -(6/6) = -1


Jadi, sumbu simetrinya adalah x = -1


Tentukan titik puncak

y0 = -(b²- 4ac/4a)

y0 = -(6²- 4x3x5/4x3)

y0 = -(36-60/12)

y0 = -(-24/12)

y0 = 2


Jadi, titik puncaknya adalah (-1, 2)

6.

OPEN ENDED QUESTION

3 mins • Ungraded

Suatu fungsi kuadrat f(x) = ax² - 4x + c mempunyai titik puncak di (1, 4). Tentukan nilai f(x)!

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Answer explanation

Pertama, substitusikan koordinat x pada titik puncak ke dalam rumus sumbu simetri untuk mendapatkan nilai a

= 1 = -(b/2a)

= 1 = -(-4/2a)

= 1 = 2/a

= a = 2


Kemudian, substitusikan nilai a dan koordinat puncak (1, 4) ke fungsi kuadrat f(x) = ax² - 6x + c untuk mendapatkan nilai c

= 1 = (2x1²) - (6x1) + c

= 1 = 2 - 6 + c

= 1 = -5 + c

= 1 + 5 = c

= 6 = c


Terakhir, untuk menemukan nilai f(x), substitusikan nilai a dan c ke dalam f(x) = ax² - 6x + c


= f(x) = ax² - 6x + c

= f(x) = 2(x²) - 6(x) + 3

= f(x) = 2x² - 6x + 3


Jadi, nilai f(x) = 2x² - 6x + 3

7.

OPEN ENDED QUESTION

3 mins • Ungraded

Suatu fungsi kuadrat f(x) = ax² - 8x + c mempunyai titik puncak di (2, 3). Tentukan nilai f(3)!

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Answer explanation

Pertama, substitusikan koordinat x pada titik puncak ke dalam rumus sumbu simetri untuk mendapatkan nilai a

= 2 = -(b/2a)

= 2 = -(-8/2a)

= 2 = 4/a

= a = 2

Kemudian, substitusikan nilai a dan koordinat puncak (2, 3) ke fungsi kuadrat f(x) = ax² - 8x + c untuk mendapatkan nilai c

= 2 = (2x2²) - (8x2) + c

= 2 = 8 - 16 + c

= 2 = -8 + c

= 10 = c

= 10 = c

Terakhir, untuk menemukan nilai f(3), substitusikan x = 3, nilai a dan c ke dalam f(x) = ax² - 8x + c

= f(x) = ax² - 8x + c

= f(3) = (2x3²) - (8x3) + 10

= f(3) = 18 - 24 + 10

= f(3) = 4

Jadi, nilai f(3) adalah 4

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