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Write a quadratic function in standard form whose graph satisfies the given conditions.
passes through
0, 0 , 2, 4 , and 4, 0

Sagot :

Mhanifaidn

Answer:

  • y = -x² + 4x

Step-by-step explanation:

The standard form:

  • y = ax² + bx + c

Pairs given:

  • (0, 0), (2, 4), (4, 0)

Substitute values of x and y and work out the values of a, b and c.

1. Point (0, 0)

  • 0 = a(0)² + b(0) + c  ⇒ c = 0

2. Point (2, 4)

  • 4 = a(2)² + b(2) ⇒ 4a + 2b = 4 ⇒ 2a + b = 2 ⇒ b = 2 - 2a

3. Point (4, 0)

  • 0 = a(4)² + b(4) ⇒ 16a + 4b = 0

Substitute b and solve for a:

  • 16a + 4(2 - 2a) = 0
  • 16a + 8 - 8a = 0
  • 8a = -8
  • a = -1

Find b:

  • b = 2 - 2(-1) = 2 + 2 = 4

The function is:

  • y = -x² + 4x

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