>>12192787>cont'dIf you have ever heard anyone say "find the roots of a polynomial", that means find the x coordinate for which the y coordinate is 0, aka, where the graph intersects the x axis (which lies at y=0). If you have the factors of a polynomial, then you know its roots, because you know that
>y=x^2+12x+36, so>y=(x+6)(x+6)You want the x for which y = 0, so you write 0 in place of y as
>0 = (x+6)(x+6)Then you can just divide out one of the factors and simply have 0 = x+ 6, so x=-6 is the x intercept, and since its the same for both factors, its the only x intercept.
One thing that's very interesting about polynomials is that if they have a root, they have it in the factor (so it goes both ways, roots find factors and factors find roots). The reason it goes backwards is a bit harder to prove, but the idea is that if you have a root labeled "r" then you can turn any polynomial p into (x-r)*q where q is a different polynomial, and since (x-r)*q is still p, the roots of p are still in q (since they're not in (x-r)), so q can be factored again. The reason this is important is because for polynomials like x^2+7x+36, and x^2+1 = 0, the fact that they cant be factorized means they also have no roots, never intersect with the x axis, because if they did, you could find factors. For our polynomial (x+6)^2, if we look at its graph we see it intersects the x axis only at x= -6, meaning its only root is -6, so it must show up twice as (x-(-6))(x-(-6)). We know it shows up twice because x+6*q works, and q cant be anything other than the root -6. We also know because x+6 is order 1, x^1, but we need x^2.
Lastly, the way to find the vertex of a parabola, its lowest point, is to find its intersection points, and since its symmetric, find the middle of those two. Then using that middling x coordinate, find the related y coordinate.
