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Function (in factored form): Function (in standard form): End behavior: As x???, f(x) ? ?. As x??, f(x) ? Roots (with multiplicity): 2 (mult. 1), -3 (mult. 2), 0 (mult. 2) Value of leading coefficient: -1 Domain: Range: Graph:

          Function (in factored form):
Function (in standard form):
End behavior:
As x???, f(x) ? ?.
As x??, f(x) ?
Roots (with multiplicity):
2 (mult. 1), -3 (mult. 2), 0 (mult. 2)
Value of leading coefficient: -1
Domain:
Range:
Graph:
        
Function (in factored form):
Function (in standard form):
End behavior:
As x???, f(x) ? ?.
As x??, f(x) ?
Roots (with multiplicity):
2 (mult. 1), -3 (mult. 2), 0 (mult. 2)
Value of leading coefficient: -1
Domain:
Range:
Graph:

Added by Victoria R.

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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Function (in factored form): Function (in standard form): End behavior: As x->-infty ,f(x)->infty . As x->infty ,f(x)-> Roots (with multiplicity): 2 (mult. 1), -3 (mult. 2), 0 (mult. 2) Value of leading coefficient: -1 Domain: Range: EXPLAIN EACH STEP PLEASE Function (in factored form): Function (in standard form): End behavior: Asx-0ofx-co As x-0o,fx)- Graph: Roots (with multiplicity): 2mult.1),-3mult.2,0mult.2 Value of leading coefficient:-1 Domain: Range:
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Transcript

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00:01 So in this problem, we're being asked to find a polynomial function in factored form that satisfies all the following conditions.
00:07 So i'm actually going to start with the conditions at the bottom.
00:09 So we're told that we have one zero with an even multiplicity and one with an odd multiplicity.
00:14 So let's say that we had a zero, let's say x was equal to three, and this had an even multiplicity.
00:19 Let's say the multiplicity was two.
00:21 Then we have an odd multiplicity, or another zero, let's call it negative one, and it has an odd multiplicity.
00:27 Let's call it 3.
00:29 So if we were the right disin factored form, remember to get from your zeros to your factors, use the opposite sign.
00:34 So if 3 is a 0, we know that x minus 3 is a factor.
00:38 And if negative 1 is a 0, we know x plus 1 is a factor.
00:41 Now, the multiplicity tells us what the exponent for our factor will be.
00:45 So for x minus 3, the exponent will be 2.
00:48 And x plus 1, it'll be 3.
00:50 Now, we're also told we have two imaginary zeros.
00:53 So let's say one of our imaginary zeros was i.
00:55 Well, that would mean that its complex conjugate also has to be a zero.
00:59 So that would be negative i...
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