second degree, with zeros of  − 6 - 6 and  2 , 2 , and goes  t o − ∞ t o - ∞ as  x − ∞ x - ∞

Question

College algebra polynomial function

Answer 1

Greetings,

There is information missing in the question, more specifically a point (x,y) that lies on the curve of a second degree polynomial.

With that said, a second degree polynomial is a quadratic function when represented as function in "Factored Form" as:

y = a (x - r₁)(x - r₂), where "a" is the leading coefficient and "r₁" and "r₂" are the zeros/roots/x-intercepts on the graph.

Secondly, these can be written as odd or even multiplicities, factors that are repeatable and odd or even amount of times. Recall, that the roots/factors that are of an odd multiplicities cross over the x-axis from a positive "y" quadrant of values to a quadrant of negative "y" values, thus positive, negative or negative, positive.

Even multiplicities touch the root/x-intercept and bounce back in the range region whence it came. So if the curve of the polynomial is in the quadrants where "y" values are positive, the curve after touching the x-intercept will bounce back in the positive region, or positive, positive. If "y" values are negative, then it will be negative, negative.

Returning back to your initial question, we are told that we have a second degree polynomial with two roots of -6 and +2, thus these are double roots, a.k.a "event multiplicities." We will convert these roots back to their factored form as follows:

Step 1): Set -6 and +2 equalled to x as follows:

x = - 6 & x = +2

Step 2) Use algebra to manipulate the above expressions set equal to zero to return them to factored form

x + 6 = 0 & x - 2 = 0 → Putting parentheses, we get (x + 6) and (x - 2)

***NOTE*** It is unclear from your initial question if this should be a degree two polynomial, for if the solutions are doubled or duplicated it will be a degree four instead***

Step 3) Since both roots are even multiplicities based on the initial question, we place a power of two to indicate a multiplicity of two, since these factors duplicate each other twice.

(x + 6)² and (x - 2)²

Step 4): Create the degree polynomial using the the above factored form template as follows:

Possible Solution 1:

y = a (x + 6)²(x - 2)², thus this would be a degree four polynomial with roots of -6 and 2 both having multiplicities of two because they occur/repeat twice. We can find the value of a, if we are given a point (x,y) that lies on the curve of this function

Possible Solution 2:

y = a (x + 6)(x - 2), should we need to infer from your initial question that -6 and 2 repeat once and not twice, they would have odd multiplicities, or powers of 1, in which 1 is understood and is not notated by understood. Thus, this solution creates a degree two polynomial, but again we need a point that lies on the curve to find the true leading coefficient a, to represent the actual polynomial.