The axis of symmetry for the graph of the function f(x) = 3x^2 + bx + 4 is x = 3/2 . What is the value of b?

b = -9

Step-by-step explanation:

Axis of symmetry of a parabola = -b / 2a.

Where y = ax² + bx + c.

Given f(x) = 3x^2 + bx + 4,

a = 3, and c = 4.

If the axis of symmetry is 3/2, than b can be found by substitution and algebra.

[ x = -b / 2a ] → [ 3 / 2 = -b / 2a ] →

[ 3 / 2 = -b / 2 (3) ] → [ 3 / 2 = -b / 6 ] →

[ 3 / 2 × 6 = -b / 6 × 6 ] → [ 18 / 2 = -b ] →

[ 9 = -b ] → [ -b = 9 ] → [ (-)(-b) = (-)(9) ] →

b = -9

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph