+0

+1
104
3
+81

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)

Nov 4, 2019

#1
+106539
+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

Nov 4, 2019
#2
+106539
+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

Nov 4, 2019
#3
+106539
+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

Nov 4, 2019
edited by CPhill  Nov 4, 2019