Question

When an archer shoots an arrow, the arrow follows a parabolic path through the

air. That parabola can be represented by a quadratic function, where x represents
the arrow's horizontal distance from the archer, and y represents the arrow's height
above the ground.
1. Philip releases an arrow from a shoulder height of 4 t
The arrow reaches a maximum height of 40 ft before
it begins to fall it lands on the ground about 160 ft
away from where he is standing. Philip models this
situation with a quadratic function. What is a zero of
the function? What is the yintercept of the graph
of the function? Construct a rough graph of the situation based on your
findings, where y represents the vertical height, in feet, and x represents the
horizontal distance, in feet.
2. Philip shoots an arrow from the top of a hill onto the ground. The flight of the
arrow can be modeled with the quadratic function y=-52² +15x+90, where
y represents the vertical height of the arrow, in meters, and x represents the
time, in seconds. Use factoring to find the positive zero of the function. What
does this zero represent? Then find the maximum value of the function by
completing the square. What does the maximum value represent? Explain
each step in your answer.

Answer 1

a

Explanation:

b

Answer 2

1. Arrow path: Quadratic function "y = -ax^2 + 40" with zeros at "± √(40/a)" and y-intercept at 40.

2. Arrow on hill: Quadratic "y = -52x^2 + 15x + 90," positive zero and max value "7275/13."

1. Quadratic Function for Arrow Path:

Given the quadratic function "y = ax^2 + bx + c" representing the arrow's path, where "a" is a constant, the arrow reaches a maximum height of 40 ft and lands 160 ft away. With "y = -ax^2 + 40," let's find the zeros (x-intercepts) and y-intercept.

Zeros (x-intercepts):

Set "y" to 0 and solve for "x":

"0 = -ax^2 + 40"

Solving for "x," we get "x = ± √(40/a)". This represents the horizontal distance where the arrow hits the ground.

Y-intercept:

When "x = 0," "y = 40". Thus, the y-intercept is 40, providing the initial height.

2. Quadratic Function for Arrow on Hill:

The quadratic function is "y = -52x^2 + 15x + 90". Let's find the positive zero and maximum value.

Positive Zero:

Set "y" to 0 and solve for "x":

"0 = -52x^2 + 15x + 90"

Factoring or using the quadratic formula provides the positive zero, representing the time the arrow hits the ground.

Maximum Value:

Complete the square to express the function in vertex form ("y = a(x - h)^2 + k"):

"y = -52x^2 + 15x + 90"

"y = -52(x^2 - 15/52x) + 90"

"y = -52(x^2 - 15/52x + (15/104)^2) + 90 + 52(15/104)^2"

Simplifying gives "y = -52(x - 15/104)^2 + 7275/13," where the maximum value is "7275/13."