Hi, I need help finding the domain and range of the Quadratic Model on part b. I've included part a. just in case you may need it.

Question

a.The parabolic path of a thrown ball can be modeled by the table. the top of the wall is at (5,6). Will the ball go over the wall? If not,will it hit wall on the way up, or the way down?

b. what is the reasonable domain and range for the function that models the path of the ball?

X:1,2,3
Y:3,5,6

How do I find the domain and range not only in this but other problems too? My book says the answer is: Domain:0 is less than or equal to x, x is less than or equal to 7. Range: 0 is less than or equal to y,y is less than or equal to 6 1/8.

Answer 1

y = a + b x^2
3 = a + b (1)^2 = a + b
5 = a + b(2)^2 = a + 4 b

b = 3 - a
5 = a + 4(3-a)
5 = a + 12 - 4a
3 a = 7
a = 7/3
b = 9/3 - 7/3 = 2/3
so
y =7/3 +2/3 x^2
check last point
y = 7/3 + (2/3)(9) = 7 1/3

I do not think your table is a parabola

Answer 2

well, maybe
y = a + b x + c x^2

3 = a + b(1) + c(1)
5 = a + b(2) + c(4)
6 = a + b(3) + c(9)
so
2 = b + 3 c
1 = b + 5 c
----------
1 = -2c
c = -1/2

1 = b -5/2
b = 7/2

3 = a + 7/2 -1/2 = a + 3 so a = 0

y = 0 + 7 x/2 - x^2/2

y = (7/2) x - (1/2)x^2
or
2 y = 7 x - x^2

Answer 3

y = (7/2) x - (1/2)x^2

reasonable domain is above ground, y>0 so
solve for zeros of x
x^2 -7x = 0
x = 0 to x = 7
range (vertex is when x = half 7 or3.5
ymax = (7/2)(7/2) -(1/2)(7/2)^2
= (1/2)(49/4) = 49/8 = 6.125
so range is 0 to 6.125

Answer 4

Now that point (5,6)
5 is beyond x = 3.5 so on the way down
y = (7/2)(5)- (1/2)(25)
= 10/2 = 5
will not make it over wall