SOLUTION: I have an algebra 2 question that is about finding an polynomial function that will create three "hills" on a graph. These hills can't be crazy numbers and have to be in the first

Question 1042006: I have an algebra 2 question that is about finding an polynomial function that will create three "hills" on a graph. These hills can't be crazy numbers and have to be in the first quadrant. The X-axis is the ground. I asked my teacher and she said that I should compress x^6 and it should work out. However, whenever I graph x^6, it always ends up being a parabola. How can I find the correct equation? Thank you so much for the time and effort :)
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
x^6 is not a parabola by itself. No compression factor alone will put all of the "hills" of x^6 in the first quadrant. You would want maybe six binomials multiplied, each with an appropriate root which will put x-intercepts to the right of the origin, and then ADD a positive constant of such value that all maxima are ABOVE the x-axis.

You learn about this topic in College Algebra, or something maybe in highschool more advanced than "Intermediate" Algebra.

Let a, b, c, d, e, be positive real numbers.

----- no hills, just a line.
------- again, no hills; just a line.

-------- degree two function, would become polynomial; and ONE hill.

-------degree three function, would become polynomial when multiplied; and ONE hill and ONE valley.

-------this would become polynomial of degree four. Two hills and ONE valleys.

-------polynomial degree five. Two hills and two valleys.

What comes next?
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