Which of the following is a tactful response to give when a student misbehaves?
2 answers:
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Answer:
(1, 3)
Step-by-step explanation:
You are given the h coordinate of the vertex as 1, but in order to find the k coordinate, you have to complete the square on the parabola. The first few steps are as follows. Set the parabola equal to 0 so you can solve for the vertex. Separate the x terms from the constant by moving the constant to the other side of the equals sign. The coefficient HAS to be a +1 (ours is a -2 so we have to factor it out). Let's start there. The first 2 steps result in this polynomial:
. Now we factor out the -2:
. Now we complete the square. This process is to take half the linear term, square it, and add it to both sides. Our linear term is 2x. Half of 2 is 1, and 1 squared is 1. We add 1 into the set of parenthesis. But we actually added into the parenthesis is +1(-2). The -2 out front is a multiplier and we cannot ignore it. Adding in to both sides looks like this:
. Simplifying gives us this:
On the left we have created a perfect square binomial which reflects the h coordinate of the vertex. Stating this binomial and moving the -3 over by addition and setting the polynomial equal to y:
From this form,
you can determine the coordinates of the vertex to be (1, 3)
Assuming you want the cosine series expansion over an arbitrary symmetric interval ![[-L,L]](https://tex.z-dn.net/?f=%5B-L%2CL%5D)
, 
, the cosine series is given by
You have
Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of
and so
So the cosine series for
periodic over an interval is
You have
Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of
and so
So the cosine series for