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3 years ago

The graph above represents the following polynomial function.

Mathematics

1 answer:

Korvikt [17]3 years ago

6 0

Answer:

C. 0, -1, 3

Step-by-step explanation:

Note: There is no graph. I assume you mean "zeroes/roots" by "solutions".

x(x^2-2x-3)=0

/(x^2-2x-3)

x=0

x^2-2x-3 = x^2+x-3x-3 = x(x+1) + ( -3(x+1) = (x-3)(x+1)

/(x-3) --> x +1 =0, x=-1

/(x+1) --> x-3 = 0, x=3

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kolezko [41]

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3 years ago

Read 2 more answers

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klemol [59]

Answer:

See proof below

Step-by-step explanation:

Assume that V is a vector space over the field F (take F=R,C if you prefer).

Let $x\in span(S_1)$ . Then, we can write x as a linear combination of elements of s1, that is, there exist $v_1,v_2,\cdots,v_k \in S_1$ and $a_1,a_2,\cdots,a_k\in F$ such that $x=a_1v_1+a_2v_2+\cdots+a_kv_k$ . Now, $S_1\subseteq S_2$ then for all $y\in S_1$ we have that $y\in S_2$ . In particular, taking $y=v_j$ with $j=1,2,\cdots,k$ we have that $v_j\in S_2$ . Then, x is a linear combination of vectors in S2, therefore $x\in span(S_2)$ . We conclude that $span(S_1)\subseteq span(S_2)$ .

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