I need help plzz.

Dilate this by 1/3

For this shape there are four points. The plotted ordered pairs are:
M = (3, 9)
N = (6, 0)
O = (3, 6)
P = (-3, 0)

All you have to do is multiply the BOTH x- and y-values by 1/3 on EACH ordered pair. And if it’s easier, you can divide by 3 as well. It’s the same thing.

M = (3, 9) = (3*1/3, 9*1/3) = (1, 3)
N = (6, 0) = (6*1/3, 0*1/3) = (2, 0)
O = (3, 6) = (3*1/3, 6*1/3) = (1, 2)
P = (-3, 0) = (-3*1/3, 0*1/3) = (-1, 0)

(1, 3) , (2, 0) , (1, 2) , and (-1, 0) are your points.

He is right math is not really my strong suit

Answer: f(x) = -2x^3 +3x^2 +11x -6see attachedan infinite number. since the magnitude of the leading coefficient is not specified, it may be any negative number. (we have chosen the smallest magnitude integer that makes all coefficients be integers.)step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.

I  dont  know much about vertexes but i know to solve this function you need to use foil method and then corresponmd the vertex with the actual coordinates, hope i !