The students in mr. wilson's physics class are making golf ball catapults. the
flight of group a's ball is modeled by the equation y = -0.014x^2 + 0.68x, where x is the ball's
distance from the catapult. the units are in feet.

a. how far did the ball fly?

b. how high above the ground did the ball fly?

c. what is a reasonable domain and range for this function?

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is and the range is

Step-by-step explanation:

we have

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's  distance from the catapult in feet

y is the flight of the balls in feet

Part a)  How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

is equal to

in this problem we have

so

substitute in the formula

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

Find out the derivative and equate to zero

Solve for x

Alternative method

To determine the x-coordinate of the vertex, find out the midpoint  between the x-intercepts

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A  reasonable domain is the distance between the two x-intercepts

so

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A  reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!