If v is the vector space from r to r, is the set of polynomial functions a subset of v? (i know i have to show the set of polynomials isn't empty, and is closed under addition and scalar multiplication. but with it being generic, with no functions given, i am a little stuck)

Here I will assume that is the vector space of functions . If is the space of continuous functions, there will be no changes in the proof, because polynomial functions are also continuous.

Let us write for the set of polynomials with real coefficients. As we want to prove that is a subspace of we need to check three conditions:

First: is a non empty set, in particular . This is not difficult, because the function identically 0 is a polynomial. Then, is not empty and contains the 0 vector of .

Second: The addition of two elements of stays in . Consider two arbitrary polynomials and . Then,

.

This means that is a polynomial of degree with coefficients , and is a polynomial of degree with coefficients . With out lose of generality assume that .

Now, notice that

which is a polynomial too.

Hence, .

Third: The multiplication by a scalar stays in .

Consider an arbitrary polynomial

and a real number .

Then,

which is a polynomial.

Hence, .

Therefore, is a subspace of the vector space .

step-by-step explanation:

c: y > - 1/6x + 1

step-by-step explanation:

dashed line - meaning this < or > also not equal

solid line - meaning this < or > also equal

your answer is c because y > means up & it's shaded up, but y < means down

hope this ! : )