Hey, I need help with some of the following problems:

1. The height (in meters) of a shot cannonball follows a trajectory given by h(t) = -4.9t^2+14t-0.4 at time t (in seconds). For how many seconds is the height of the cannonball at least 6 meters?

(I think that may be answered before but I am still a bit confused)

2.Find the maximum value of f(x)=-4(x-1)^2+6.

(I'm getting (1,6) as vertex and idk why its wrong)

3. What is the maximum value of 4(x+7)(2-x) over all real numbers ?

(again not sure why im wrong)

doorknoob Nov 4, 2019

#1**+1 **

1. The height (in meters) of a shot cannonball follows a trajectory given by h(t) = -4.9t^2+14t-0.4 at time t (in seconds). For how many seconds is the height of the cannonball at least 6 meters?

We want to solve this

-4.9t^2 + 14t - 0.4 = 6 subtract 6 from both sides

-4.9t^2 + 14t - 6.4 = 0 multiply through by -10

49t^2 - 140t + 64 = 0 factor as

(7 t - 16) ( 7t - 4) = 0

Set each factor to 0 and solve for t and it will be at 6m at 4/7 sec and 16/7 sec

So....it will be a height of 6m (or above) for 16/7 - 4/7 = 12/ 7 seconds

CPhill Nov 4, 2019

#2**+1 **

2.Find the maximum value of f(x) = -4(x-1)^2+6.

The vertex is (1 , 6)

So....the max value is 6

You are correct....this graph proves it : https://www.desmos.com/calculator/ewjtfy8u96

CPhill Nov 4, 2019

#3**+1 **

3. What is the maximum value of 4(x+7)(2-x) over all real numbers ?

4 [ 2x + 14 - x^2 - 7x ] =

4 [ -x^2 - 5x + 14]

-4x^2 - 20x + 56

The x coordinate of the vertex = - (-20) / [ 2 * -4] = 20 / -8 = -5/2 = -2.5

So.....the max is

-4 (-5/2)^2 - 20 (-5/2) + 56 =

-4 ( 25/4) + 50 + 56 =

-25 + 50 + 56 =

81

Here's the graph : https://www.desmos.com/calculator/pzjfnx8ugd

CPhill Nov 4, 2019