Question

Quadratic models question

Answer 1

I don't know that!! Sorry

Answer 2

@ranga could you do this?

Answer 3

f(t) = -16t^2 + 20t
complete the square.
-16(t^2 - 20/16t) = -16(t^2 - 5/4t) = -16{ (t-5/8)^2 - (5/8)^2 } =
-16{ (t-5/8)^2 - 25/64 } = -16(t-5/8)^2 + 25/4
The vertex is at (5/8, 25/4).
So the maximum of f(t) is 25/4 = 6.25 feet.

For g(t) you can look at the table and see that the maximum value is at 9 feet.

So the second dog, the foxhound, jumps higher (9 feet) than the first dog, the Labrador (6.25 feet).

Answer 4

B) f(t) = -16t^2 + 20t = -4t(4t - 5)
To find x-intercept, set f(t) = 0 and solve for t:
-4t(4t-5) = 0
t = 0 and t = 5/4 = 1.25 seconds

For g(t) we can see the x-intercepts are at:
t = 0 and t = 1.5 seconds.

The foxhound has a higher x-intercept at 1.5 seconds than the Labrador at 1.25 seconds.

What do the x-intercepts represent?
They represent the time t at which the dogs are on the ground. At t = 0 they are both on the ground. Then they leap up and are off the ground. A little later they land back on the ground. The Labrador is back on the ground after 1.25 seconds and the foxhound is back on the ground after 1.5 seconds. A higher x-intercept means the foxhound stays in the air a little longer than the Labrador.

Answer 5

C) To find y-intercept, set t = 0 and find f(t) from the equation and g(t) from the table.
f(t) = -16t^2 + 20t
put t = 0
f(0) = 0

From the table, when t= 0, g(0) = 0.

It means, at time t = 0, f(0) = 0 and g(0) = 0. That is both dogs are on the ground. They are not at an elevation or on a table or at some height when t = 0. They are both on the ground.