For this case, the main function is given by:
We can apply the following transformation:
Horizontal displacements:
Assume h> 0:
To graph f (x + h) move the graph of f (x) h units to the left.
For h = 1 we have:
Answer:
the graph y=6(x+1)^2 is the graph y=6x^2 moved 1 unit to the left.
The function you seek to minimize is
()=3‾√4(3)2+(13−4)2
f
(
x
)
=
3
4
(
x
3
)
2
+
(
13
−
x
4
)
2
Then
′()=3‾√18−13−8=(3‾√18+18)−138
f
′
(
x
)
=
3
x
18
−
13
−
x
8
=
(
3
18
+
1
8
)
x
−
13
8
Note that ″()>0
f
″
(
x
)
>
0
so that the critical point at ′()=0
f
′
(
x
)
=
0
will be a minimum. The critical point is at
=1179+43‾√≈7.345m
x
=
117
9
+
4
3
≈
7.345
m
So that the amount used for the square will be 13−
13
−
x
, or
13−=524+33‾√≈5.655m
Answer:
As we know he pinched 15 games and he pitched 3/5, we can do a rule of three:
15 - 3/5
x 1
We out a 1 ad that's the total, 5//5
So now we know x = , = 25
He pitched 25 games