-the graph of y=f(x) extends down into quadrant lll and down into quadrant IV
-the zeros of f are -2,3 and 5
-the degree of f<5
1.what degrees is possible for f?
2.what are the maximum and minimum multiplicities possible for a zero of f?
3.two possible equations?
4.two possible graphs of y=f(x)
4 years ago
Answered By Vincent E
Hello! First let's sketch what we know. The two arms of the function are both going down. What can we say about some characteristics of the function? Since the arms are both going in the same direction, we know that the function has an even degree. Also, since the right arm is going down, the function will have a negative leading coefficient. Now that we've established those facts let's answer the questions.
1. What degrees are possible for f? We've just determined in our initial analysis that the degree of the function needs to be even. In the blurb we are told that the degree is less than 5. Therefore the only possible degrees for f are 2 and 4.
2. What are the maximum and minimum multiplicities possible for a zero of f? We know that there are 3 possible multiplicities: 1 (straight line through x axis), 2 (parabola touching the x axis), and 3 (inflection point). We also know that the multiplicities of the zeros of f have to add up to the degree of f. We know from Q1 that the degree is either 2 or 4. So let's look:
- For degree 2: we can have 2 zeros of multiplicity 1.
- For degree 4: we can have 2 zeros with both multiplicity 2 OR 1 zero with multiplicity 1 and 1 zero with multiplicity 3.
Therefore, the minimum multiplicity is 1 and the maximum is 3.
3. Two possible equations? Let's use the info from question 1 and 2.
- For degree 2, we said that we can have 2 zeros of multiplicity 1. From the blurb we are told there are zeroes at x=-2.3 and x=5. We also know that f has a negative leading coefficient. Therefore:
$f\left(x\right)=-\left(x+2.3\right)\left(x-5\right)$ƒ(x)=−(x+2.3)(x−5). You can put the 2.3 into fractions if you want.
- For degree 4, we said that we can have either 2 zeros with both multiplicity 2 OR 1 zero with multiplicity 1 and 1 zero with multiplicity 3. Lets do the 2nd case. From the blurb we are told there are zeroes at x=-2.3 and x=5. We also know that f has a negative leading coefficient. Therefore:
$f\left(x\right)=-\left(x+2.3\right)\left(x-5\right)^3$ƒ(x)=−(x+2.3)(x−5)3. You can put the 2.3 into fractions if you want. The multiplicity 3 can either be assigned to -2.3 or 5 (but not both).
4. Graph the 2 equations from above and confirm the information from the blurb is correct (eg. left arm down into Q3, right arm down into Q4, zeros at -2.3 and 5, etc.).
4 years ago
Answered By Vincent E
Hello! First let's sketch what we know. The two arms of the function are both going down. What can we say about some characteristics of the function? Since the arms are both going in the same direction, we know that the function has an even degree. Also, since the right arm is going down, the function will have a negative leading coefficient. Now that we've established those facts let's answer the questions.
1. What degrees are possible for f? We've just determined in our initial analysis that the degree of the function needs to be even. In the blurb we are told that the degree is less than 5. Therefore the only possible degrees for f are 2 and 4.
2. What are the maximum and minimum multiplicities possible for a zero of f? We know that there are 3 possible multiplicities: 1 (straight line through x axis), 2 (parabola touching the x axis), and 3 (inflection point). We also know that the multiplicities of the zeros of f have to add up to the degree of f. We know from Q1 that the degree is either 2 or 4. So let's look:
- For degree 2: we can have 2 zeros of multiplicity 1.
- For degree 4: we can have 2 zeros with both multiplicity 2 OR 1 zero with multiplicity 1 and 1 zero with multiplicity 3.
Therefore, the minimum multiplicity is 1 and the maximum is 3.
3. Two possible equations? Let's use the info from question 1 and 2.
- For degree 2, we said that we can have 2 zeros of multiplicity 1. From the blurb we are told there are zeroes at x=-2.3 and x=5. We also know that f has a negative leading coefficient. Therefore:
$f\left(x\right)=-\left(x+2.3\right)\left(x-5\right)$ƒ (x)=−(x+2.3)(x−5). You can put the 2.3 into fractions if you want.
- For degree 4, we said that we can have either 2 zeros with both multiplicity 2 OR 1 zero with multiplicity 1 and 1 zero with multiplicity 3. Lets do the 2nd case. From the blurb we are told there are zeroes at x=-2.3 and x=5. We also know that f has a negative leading coefficient. Therefore:
$f\left(x\right)=-\left(x+2.3\right)\left(x-5\right)^3$ƒ (x)=−(x+2.3)(x−5)3. You can put the 2.3 into fractions if you want. The multiplicity 3 can either be assigned to -2.3 or 5 (but not both).
4. Graph the 2 equations from above and confirm the information from the blurb is correct (eg. left arm down into Q3, right arm down into Q4, zeros at -2.3 and 5, etc.).
Please reach out if you need more help!
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