Consider the function represented by the equation 6q = 3s - 9. write the equation in function notation, where q is the independent variable

a. f(q) = 1/2q - 3/2

b. f(q) = 2s + 3

c. f(s) = 1/2s - 3/2

d. f(q) = 2q + 3

  D.  f(q) = 2q + 3

Step-by-step explanation:

Choices B and C are eliminated right away because there is no "q" in the definition, just "s". If the function is to be a function of "q", then "q" is expected to appear in the definition.

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The two variables in the given equation are "s" and "q". If "q" is designated as the independent variable, then the dependent variable is "s". The equation must be solved for "s":

  6q = 3s - 9

We observe that the coefficient of "s" is 3, and that all numbers are multiples of 3, so we can divide by 3 to simplify this a bit:

  2q = s - 3

Since we want an expression for s alone, we can add 3 to get ...

  2q +3 = s

Now, we can write ...

  s = f(q) = 2q +3