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kirill115 [55]
3 years ago
6

Ming throws a stone off a bridge into a river below. The stone's height (in meters above the water), x seconds after Ming threw

it, is modeled by: h(x)=−5(x−1)^2+45 What is the maximum height that the stone will reach?
Mathematics
1 answer:
mina [271]3 years ago
6 0

Answer:

The maximum height that the stone will reach is 45 meters.

Step-by-step explanation:

A quadratic function is one that can be written as an equation of the form:

f (x) = ax² + bx + c

where a, b, and c (called terms) are any real numbers and a is nonzero (can be greater or less than zero). The value of b and c itself can be zero.

So in the quadratic equation

  • ax ² is the quadratic term
  • bx is the linear term
  • c is the independent term

The vertex of a quadratic equation or parabola is the highest or lowest point of the graph corresponding to that function. The vertex is in the plane of symmetry of the parabola; anything that happens to the left of this point will be an exact reflection of what happens to the right. In other words, the vertex is a point that is part of the parabola, whose ordinate is the value  minimum or maximum function. At that point an imaginary axis can be drawn that makes  symmetric the graph of the function, which is called the axis of symmetry.

The quadratic function can be expressed using the canonical form:

f (x) = a. (x - xv) ² + yv

where (xv, yv) are the vertex coordinates.

So, in the case of h(x)=−5(x−1)²+45, the vertex is (1,45) [xv=1,yv=45]

If a> 0 (positive) the parabola is concave, that is, the parabola faces upwards.

If a <0 (negative) the parabola is convex, that is, the parabola faces downward.

If the parabola faces up, the vertex will be its minimum value. If the parabola faces downwards, the vertex will be its maximum value.

In the case of h(x)=−5(x−1)²+45, a <0 (negative) the parabola is convex  and the vertex will be its maximum value.

Being "x" the time and "y" the height , <u><em>the maximum height that the stone will reach is 45 meters.</em></u>

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Step-by-step explanation:

A = \pir² = 154

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<u></u>

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