How would I go about approaching this question?

[pope4]

I tried doing this question over and over again and I can't even begin to imagine how to solve it. I wrote down the available data (x-intercepts, constant term, even or
odd exponent) and I'm still not sure how to do it. Thank you for the help!

 

[Dr.Peterson]

I tried doing this question over and over again and I can't even begin to imagine how to solve it. I wrote down the available data (x-intercepts, constant term, even or
odd exponent) and I'm still not sure how to do it. Thank you for the help!


View attachment 34297

Please show the data you wrote down, so we can be sure you found everything you need.

Here's the graph:

​

One thing you need is the x-intercepts. What are they?

Another is the multiplicities of the x-intercepts. What do you think they might be?

With those, you can make your first attempt at a polynomial, in factored form. What will that be?

Then there may be some adjustments needed (to get the right y-intercept), but that can wait.

 

[pope4]

Please show the data you wrote down, so we can be sure you found everything you need.

Here's the graph:

View attachment 34298​

One thing you need is the x-intercepts. What are they?

Another is the multiplicities of the x-intercepts. What do you think they might be?

With those, you can make your first attempt at a polynomial, in factored form. What will that be?

Then there may be some adjustments needed (to get the right y-intercept), but that can wait.

I wrote down that the x-intercepts are 1 and -2, both with multiplicities of 3, that the y-intercept is -2, therefore the constant term is -2, and that the exponent has to be positive. This led me to f(x)=a(x-1)^3(x+2)^3, but needless to say this did lead me towards a sensible solution.

 

[Dr.Peterson]

I wrote down that the x-intercepts are 1 and -2, both with multiplicities of 3, that the y-intercept is -2, therefore the constant term is -2, and that the exponent has to be positive. This led me to f(x)=a(x-1)^3(x+2)^3, but needless to say this did lead me towards a sensible solution.

I take it you did determine a, and solve the problem? The word "but" could suggest otherwise.

It may be worth noting that in factored form, there is no "constant term"; that's the y-intercept, which you use to find a by determining f(0).

I'm not sure what you mean by "the exponent has to be positive". There are no negative exponents in a polynomial. Maybe you meant even.

 

[pope4]

I take it you did determine a, and solve the problem? The word "but" could suggest otherwise.

It may be worth noting that in factored form, there is no "constant term"; that's the y-intercept, which you use to find a by determining f(0).

I'm not sure what you mean by "the exponent has to be positive". There are no negative exponents in a polynomial. Maybe you meant even.

Sorry I made a bunch of grammar mistakes in that sentence! I didn't end up finding a solution, and 'even exponent' is what I meant to say (English isn't my first language and despite quite a bit of experience I have with it I still keep mistaking 'even' and 'odd' for 'positive' and 'negative', quite a troublesome trait to have).

I wasn't sure if I was doing it right by writing down the factored solution I mentioned earlier since, after expanding it, I got some obscene equation (x^6+3x^5-3x^4-11x^3+6x^2+12x-8). However, assuming the equation was somehow correct, how would I go about finding a?

 

[blamocur]

Sorry I made a bunch of grammar mistakes in that sentence! I didn't end up finding a solution, and 'even exponent' is what I meant to say (English isn't my first language and despite quite a bit of experience I have with it I still keep mistaking 'even' and 'odd' for 'positive' and 'negative', quite a troublesome trait to have).

I wasn't sure if I was doing it right by writing down the factored solution I mentioned earlier since, after expanding it, I got some obscene equation (x^6+3x^5-3x^4-11x^3+6x^2+12x-8). However, assuming the equation was somehow correct, how would I go about finding a?

You've done a good job! All that remains is finding 'a' : what is the value of f(0) ?

 

[Steven G]

I wrote down that the x-intercepts are 1 and -2, both with multiplicities of 3, that the y-intercept is -2, therefore the constant term is -2, and that the exponent has to be even. This led me to f(x)=a(x-1)^3(x+2)^3, but needless to say this did lead me towards a sensible solution.

I am sorry but 3 is not even. What did you mean by an even exponent?

 

[Dr.Peterson]

I still keep mistaking 'even' and 'odd' for 'positive' and 'negative', quite a troublesome trait to have

I occasionally do that myself, in a context where both ideas occur (such as even powers of negative numbers).

I wasn't sure if I was doing it right by writing down the factored solution I mentioned earlier since, after expanding it, I got some obscene equation (x^6+3x^5-3x^4-11x^3+6x^2+12x-8). However, assuming the equation was somehow correct, how would I go about finding a?

You were told to keep it in factored form, for a reason. That's the natural form for this sort of problem.

You know that f(x) = a(x-1)^3(x+2)^3, and that f(0) = -2. Put those together: a(0-1)^3(0+2)^3 = -2, and solve for a. You were so close I assumed you had taken this last step. As I'd said, "Then there may be some adjustments needed (to get the right y-intercept), but that can wait."

I am sorry but 3 is not even. What did you mean by an even exponent?

That refers to the total power: This is a 6th degree equation; you can see it's even because it rises at both ends. Surely you can see that?

 

[Steven G]

Good point!