Review the graph of function g(x). On a coordinate plane, a curve starts at closed circle (4, negative 4), curves up to (3, negative 3), and curves down to (negative 1, negative 8), and then curves up through (negative 3, 4). A curve starts at open circle (4, 2) and curves up through (8, 5).

What are Limit of g (x) as x approaches 4 negative and Limit of g (x) as x approaches 4 plus, if they exist?

Limit of g (x) = 2 as x approaches 4 negative and limit of g (x) = negative 4 as x approaches 4 plus
Limit of g (x) = negative 4 as x approaches 4 negative and limit of g (x) = negative 2 as x approaches 4 plus
Limit of g (x) = negative 4 as x approaches 4 negative and limit of g (x) D N E as x approaches 4 plus
Limit of g (x) D N E as x approaches 4 negative and limit of g (x) = negative 4 as x approaches 4 plus

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step-by-step explanation:

2

step-by-step explanation:

3

step-by-step explanation:

because where the point where the 2 lines intersect is (3,-1) the x value is 3

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.