Question

Hello, I am a student taking precalculus this is the problem that I need help with

Answer 1

Hello there. To solve this question, we have to remember how to determine the equation for the polynomial function portrayed in the graph.

Given the following graph:

We know that all its roots are integers and we can see that the x-axis is uniformly spaced by the unit factor.

Hence we determine that its roots are -1 and 5.

But notice this function has the shape of a cubic function (more specifically an odd degree polynomial, but the question asks to use the smaller degree possible).

The thing is that the function only has two roots, when it should have three.

This means that one of the roots have multiplicity of two.

In this case, say we have r1, r2 and r3 the roots of this function and a as its leading coefficient, then it can be written in the canonical form:

`y=a\cdot(x-r_1)\cdot(x-r_2)\cdot(x-r_3)_`

Using 1 as its leading coefficient and say that the root with multiplicity is -1. Then it can be written as

`y=(x+1)^2\cdot(x-5)`

But we cannot do this assumption without a reason.

First, remember that the linear factor for a cubic polynomial:

`y=a_3x^3+a_2x^2+a_1x+a_0`

Is given by Vieta's formula:

`a_0=-(r_1\cdot\,r_2\cdot\,r_3)/(a_3)`

That is, the negative of the product of the roots divided by the leading coefficient.

Notice, from the graph, that its linear factor is also the y-intercept and it happens above the x-axis (meaning it is positive).

Assuming that a3 as the leading coefficient is equal to 1, then we have to get a negative product of the roots.

Means that only one root can be negative and it happens to be negative one.

Hence choosing the leading coefficient 1, the correct canonical form is:

`y=(x+1)\cdot(x+5)^2`

If we instead chose the leading coefficient to be -1, then we should have two negative roots to get a positive value and then the first answer we got would be correct:

`y=-(x+1)^2\cdot(x-5)`

Up to the minus sign in front of it. Both of these answers are correct now.