The black graph is the graph of y=f(x). choose the equation for the red graph.
!

B

Step-by-step explanation:

The red graph is a horizontal translation of 5 units left followed by a reflection in the x- axis.

Given f(x) then f(x + a) is a horizontal translation of f(x)

• If a > 0 then a shift to the left of a units

• If a < 0 then a shift to the right of a units

The black graph is shifted 5 units left, thus

f(x) → f(x + 5)

Under a reflection in the x- axis

a point (x, y ) → (x, - y )

Note that the y- coordinate of the image is the negative of the original

Note also that = - y, thus

= f(x + 5) → B

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

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.