The axis of symmetry for the graph of the function f(x) = 3x² + bx + 4 is x equals three-halves. What is the value of b?

Solution:

It is given that

f(x) = 3x² + bx + 4

a = 3, b = b and c = 4

Where x = 3/2 is the axis of symmetry.

In order to determine the x value of vertex ax² + bx + c

x value = -b/2a

3x² + bx + 4 = 0

x = 3/2

So we get

-b/2(3) = 3/2

-b/6 = 3/2

By further calculation

-b = 3/2 × 6

-b = 9

b = -9

Therefore, the value of b is -9.

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The axis of symmetry for the graph of the function f(x) = 3x² + bx + 4 is x equals three-halves. What is the value of b?

Summary:

The axis of symmetry for the graph of the function f(x) = 3x² + bx + 4 is x equals three-halves. The value of b is -9.