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Nelo
Sep18-11, 09:11 PM
1. The problem statement, all variables and given/known data


Why is this not a polynomial function? h(x) = 3x + 2x
2. Relevant equations



3. The attempt at a solution

3x+2x = 5x.

5x is a linear function with a degree of 1, why is this not a polynomial funct?

QuarkCharmer
Sep18-11, 09:25 PM
You already said it has a degree of one and is linear right?

What is it called when the term has one degree?

Nelo
Sep18-11, 09:42 PM
so.. linear functions are not polynomial functions but quadratic functions of x^2 are ?

symbolipoint
Sep18-11, 10:13 PM
An Algebra book I've seen recently would classify that as a polynomial, as well as being a monomial IF you see that you can combine the terms: 3x+2x=5x. Yes, 5x would still be a polynomial (but I would not want to call it that. I would rather just call it a monomial).

Nelo
Sep18-11, 10:26 PM
Quick question.

Describe transformation to graph x^4 :: 5f[2/5(x-3)] +1

so.. vertical 5, horizontal 2/5 (in the book it says 5/2... ? is that how it is?) right 3 up 1.

Is the horizontal 2/5 or 5/2 ? why flip it if its outside the x already?\

Or do you always state the recipricol of it?

Mark44
Sep19-11, 03:58 PM
Quick question.

Describe transformation to graph x^4 :: 5f[2/5(x-3)] +1

so.. vertical 5, horizontal 2/5 (in the book it says 5/2... ? is that how it is?) right 3 up 1.

Is the horizontal 2/5 or 5/2 ? why flip it if its outside the x already?\

Or do you always state the recipricol of it?
For new questions, you really should start a new thread.

Assuming f(x) = x4, the graph you're asking about is y = 5f( 2/5 *(x - 3)] + 1.

If you know the graph of y = g(x), the graph of y = g(3x) represents a compression toward the y-axis by a factor of 1/3 of the graph of g. So for example, if (6, 2) is a point on the graph of g, then (2, 2) will be on the graph of y = g(3x).

The graph of y = 2g(x) represents a stretch away from the x-axis by a factor of 2.

Can you apply these ideas to your problem?