Can y’all help me on num 3
2 answers:
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Part A
Given that
Then,
For
, then
Thus,
For
, we have
Part B
Recall that from part A,
Now, at initial position, t = 0 and
, thus we have
and when the velocity drops to half its value,
and 
Thus,
Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by
Given that
Then,
For
Thus,
For
Part B
Recall that from part A,
Now, at initial position, t = 0 and
and when the velocity drops to half its value,
Thus,
Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by
Answer:
(5,-6)
Step-by-step explanation:
ONE WAY:
If , then
.
Let's simplify that.
Distribute with :
Combine the end like terms :
Use identity for
:
Combine like terms and
:
We are given .
So we have that .
The vertex happens at .
Compare to
to determine
.
Let's plug it in.
So the coordinate is 5.
Let's find the corresponding coordinate by evaluating our expression named
at
:
So the ordered pair of the vertex is (5,-6).
ANOTHER WAY:
The vertex form of a quadratic is where the vertex is
.
Let's put into this form.
We are given .
We will need to complete the square.
I like to use the identity .
So If you add something in, you will have to take it out (and vice versa).
So we have in vertex form is:
.
The vertex is (3,-6).
So if we are dealing with the function .
This means we are going to move the vertex of right 2 units to figure out the vertex of
which puts us at (3+2,-6)=(5,-6).
The coordinate was not effected here because we were only moving horizontally not up/down.