Question

The parabola narrows by a factor of 3, opens down, shifts right 2 units.​

Answer 1

Answer: y=-3(x-2)^2

a(x-h)^2+k

The parabola narrows (negative) by 3 (a) and shifts right 2 (h).

Answer 2

The parabola
`\(y = -3(x - 2)^2 + k\)`models the height of a vertically launched object. The vertex, representing max height, is at (2, k), reached at 2 seconds.

Suppose an object is launched vertically, and its height,
`\(y\),`above the ground over time,
`\(x\)`, is modeled by the parabolic equation
`\(y = -3(x - 2)^2 + k\),` where
`\(y\)` is in meters and
`\(x\)` is in seconds. The given transformation indicates that the parabola narrows by a factor of 3, opens downward, and shifts right 2 units. In this context:

1. Vertex of the Parabola:

The vertex, representing the maximum height of the object, is at
`\((2, k)\).`The rightward shift by 2 units places the vertex at
`\(x = 2\),` and the downward opening is reflected in the negative coefficient,
`\(-3\),`indicating the parabola opens downward.

2. Time of Maximum Height:

The time at which the object reaches its maximum height corresponds to the
`\(x\)`-coordinate of the vertex. Therefore, the object reaches its peak at
`\(x = 2\)` seconds.

3. Initial Height:

The initial height of the object is given by the constant term \(k\). As it is not specified in the given information, the specific value of \(k\) would need to be provided to determine the initial height.

In summary, understanding the transformations applied to the original parabola allows us to analyze the object's motion, finding the vertex, time of maximum height, and, with additional information, the initial height of the launched object.

The probable question maybe:

What is the equation to model the height
`(\(y\))`of an object launched vertically, considering the parabolic form
`\(y = -3(x - 2)^2 + k\)`? Determine the vertex, time of peak height, and initial height
`(\(k\)).`