How do I write a polynomial function by looking at a graph

[chibininja22]

How do I write the equation for this graph? I know I have to find the x-intercepts but I tried solving it and graphing it to check and it didn't look like the one on the problem. This is for my Algebra II class we have a test on Friday. ?

Thanks in advance!

 

[Dr.Peterson]

Please show your attempt, so we can see what you might be doing wrong.

Perhaps you didn't fully take multiplicity into account; or perhaps you didn't take into account the vertical scale (e.g. your answer did not pass through (3, -1)), and were expected to; or you may just not have put it into the expected form.

 

[LCKurtz]

And do you see why you would expect a 5th degree polynomial for starters?

 

[chibininja22]

Please show your attempt, so we can see what you might be doing wrong.

Perhaps you didn't fully take multiplicity into account; or perhaps you didn't take into account the vertical scale (e.g. your answer did not pass through (3, -1)), and were expected to; or you may just not have put it into the expected form.

This is my work I'm not sure if it's right.

 

[chibininja22]

And do you see why you would expect a 5th degree polynomial for starters?

I think it's because one of the curves bounces off 2 so, therefore, it's going to be an even power. And the zero at 4 creates an S curve meaning that it has an odd power. I'm not sure if this is right.

 

[tkhunny]

Does f(2) give zero?

Why did you select addition to join them?

 

[Dr.Peterson]

I think it's because one of the curves bounces off 2 so, therefore, it's going to be an even power. And the zero at 4 creates an S curve meaning that it has an odd power. I'm not sure if this is right.

You have the right multiplicity; but that relates to factors, not terms. Check what your textbook or notes say about the topic; there should be examples. Or see here.

 

[LCKurtz]

I think it's because one of the curves bounces off 2 so, therefore, it's going to be an even power.

Instead of "bounces off", think double root. Or, as you say, a higher even power.

And the zero at 4 creates an S curve meaning that it has an odd power. I'm not sure if this is right.

It's close. But the odd power must be greater than 1 as a simple root would cross. Call it an inflection point instead of an S curve. You have the right idea with other corrections that have been suggested.

 

[JeffM]

I would think about this differently.

The slope of an even-powered polynomial in x is going to have opposite signs for large enough positive and negative values of x. Why? The slope of an odd-powered polynomial in x is going to have identical signs for large enough positive and negative values of x. Why?

So does this curve appear to be even-powered or odd-powered?

The maximum number of real roots that a polynomial can have equals the degree of the polynomial. A quadratic may have two real roots, but no more. A cubic may have three real roots, but no more.

So what is the lowest degree that this polynomial can have?

The maximum number of local extrema plus inflection points that a polynomial can have equals one fewer than the degree of the polynomial. Moreover, an even-powered polynomial has at least one extremum. A quadratic will have one local extremum and thus can have no inflection point. A cubic will have either no or else two local extrema, but if it has two local extrema, it will have no inflection point.

So what is the lowest degree that this polynomial can have?

Do you know the fundamental theorem of algebra? Do you know about derivatives?