1) let (f_{n}) be a sequence of real- valued functions that converges uniformly to f_{*} on a set a\subseteq r , where each f_{n} is bounded on a, i. e there exists mn> 0 such that |f_{n}(x)|\leq m_{n} for all x\in a .
a) show that f_{*} bounded on a.
b)show that there exists m> 0 shuch that |f_{n}(x)|\leq m for all x\in a and all n.
c) let (a_{n}) be a bounded real sequence , and denote by a_{n}f_{n} the product of a_{n} and f_{n} . show that (a_{n}f_{n}) has a subsequence which converges uniformly on a.

let's cover over the formula of the vertex:

y = a(x – h)^2 + k

in this formula, (h,k) are the vertex, always remember that the positive or negative value of the vertex is based on the sign it has.

for example, if the vertex is -3,-6 then both the x and y value would have the positive sign attached to it and if it were a positive x and y value, then a negative sign would be attached to it.

with that being said, for our problem the x value is a positive one so we are looking for a number with a negative sign next to it, in this case the first and second choice.

however, for the first answer choice it isn't quite part of the formula so we know that the answer choice correctly displays the x value.

to confirm, we can see if the -1 fits correctly with the second answer choice, and as mentioned above a negative number will have a positive sign attached to it. so, for our problem we can see that -1 has been changed to +1, so we can confirm that the second answer choice is correct.

hence, our final answer is y=(x - 1)^2 + 1

answer: i believe the answer is e

step-by-step explanation:

your answer   is c

step-by-step explanation:

step-by-step explanation: