Question

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

Answer 1

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)

Answer 2

Answer:

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

Step-by-step 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:

$\implies y=0.2(46.5)=9.3$