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An archer shoots an arrow up towards a target located on a hill, which is shown by the graph.

A graph of a line and a parabola intersecting at (46.5, 9.3).

Which set of equations best models the point of intersection of the arrow and the target?

y = 0.006x2 + 0.35x + 6 and y = x
y = –0.006x2 + 0.35x + 6 and y = –x
y = 0.006x2 + 0.35x + 6 and y = 0.2x
y = –0.006x2 + 0.35x + 6 and y = 0.2x

2 Answers

4 votes

Answer:


\sf y = -0.006x^2 + 0.35x + 6\quad and\quad y = 0.2x

Explanation:

As the archer shoots the arrow up, the trajectory of the arrow can be modeled as a parabola that opens downwards. Therefore, this is a quadratic equation with a negative leading coefficient.

From inspection of the answer options, the path of the arrow is :


y=-0.006x^2+0.35x+6

This means the options for the line equation are
y=-x and
y=0.2x

We are told that the point of intersection of the line and the parabola is at (46.5, 9.3). As both the x-value and y-value are positive, the equation of the line must be
y=0.2x

To confirm, substitute
x=46.5 into the equation:


\implies y=0.2(46.5)=9.3

User Joky
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3.2k points
3 votes

As no graph is shown some basic information of parabola can be helpful

  • As it's. a hill hence the parabola facing downwards
  • a is negative

And a coordinate is given check it whether it is equal to second part of options

  • (46.5,9.3)

We get

  • y≠x
  • y≠-x
  • y=0.2x

Last two are radar now

As a must be negative option D has such similarities.

  • Option D is correct (Assumption)
User DEfusion
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3.6k points