Question

Write a quadratic function in standard form whose graph satisfies the given conditions.

passes through
0, 0 , 2, 4 , and 4, 0

Answer 1

• y = -x² + 4x

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