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Zorluk: OrtaQuadratic Functions and Graphs

The trajectory of a basketball thrown toward a hoop can be modeled by a quadratic function. In the xyxy-plane, xx represents the horizontal distance in feet from the shooter and yy represents the height of the basketball in feet. The basketball reaches its maximum height of 1414 feet at a horizontal distance of 88 feet from the shooter. If the basketball is released at a height of 66 feet, which of the following equations models the trajectory of the basketball?

  1. A
    y=−18(x+8)2+14y = -\frac{1}{8}(x + 8)^2 + 14
  2. B
    y=−18(x−14)2+8y = -\frac{1}{8}(x - 14)^2 + 8
  3. y=−18(x−8)2+14y = -\frac{1}{8}(x - 8)^2 + 14Cevap
  4. D
    y=−8(x−8)2+14y = -8(x - 8)^2 + 14

Cevap

The equation y=−18(x−8)2+14y = -\frac{1}{8}(x - 8)^2 + 14
The vertex form of a quadratic function is y=a(x−h)2+ky = a(x - h)^2 + k, where (h,k)(h, k) represents the coordinates of the vertex. Since the maximum height of the basketball is 1414 feet at a horizontal distance of 88 feet, the vertex is (8,14)(8, 14). Substituting these values gives the equation y=a(x−8)2+14y = a(x - 8)^2 + 14. The release point represents the y-intercept where x=0x = 0 and y=6y = 6. Substituting these coordinates into the equation gives 6=a(0−8)2+146 = a(0 - 8)^2 + 14, which simplifies to 6=64a+146 = 64a + 14. Subtracting 1414 from both sides results in −8=64a-8 = 64a, and dividing both sides by 6464 yields a=−18a = -\frac{1}{8}. Therefore, the correct trajectory model is y=−18(x−8)2+14y = -\frac{1}{8}(x - 8)^2 + 14.

Adım Adım Çözüm

1
Identify the vertex coordinates (h,k)(h, k) of the parabolic path from the given maximum height context.
The vertex (h,k)(h, k) is (8,14)(8, 14).
The basketball reaches its maximum height of 1414 feet at a horizontal distance of 88 feet, and the maximum of a downward-opening parabola is its vertex.
2
Write the general vertex form of a quadratic equation and substitute the coordinates of the vertex.
y=a(x−8)2+14y = a(x - 8)^2 + 14
The vertex form of a quadratic equation is y=a(x−h)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex.
3
Substitute the initial release point (0,6)(0, 6) into the equation to solve for the coefficient aa.
6=a(0−8)2+14  ⟹  6=64a+14  ⟹  −8=64a  ⟹  a=−864=−186 = a(0 - 8)^2 + 14 \implies 6 = 64a + 14 \implies -8 = 64a \implies a = -\frac{8}{64} = -\frac{1}{8}
The basketball is released at a height of 66 feet when the horizontal distance x=0x = 0, representing the y-intercept of the trajectory.
4
Substitute the solved value of aa back into the vertex form equation.
y=−18(x−8)2+14y = -\frac{1}{8}(x - 8)^2 + 14
This yields the complete quadratic model representing the path of the basketball.

Anahtar Kavram

Writing quadratic equations in vertex form from context
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Konular

Quadratic Functions and Graphs

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