F(x) = 2x - 4
f(2 ≤ x) = 2(2 ≤ x) - 4
f(x ≥ 2) = 2(x ≥ 2) - 4
f(x ≥ 2) = 2(x) ≥ 2(2) - 4
f(x ≥ 2) = 2x ≥ 4 - 4
f(x ≥ 2) = 2x ≥ 0
f(x ≥ 2) = x ≥ 0
f(x) = 2x - 4
f(x ≤ 6) = 2(x ≤ 6) - 4
f(x ≤ 6) = 2(x) ≤ 2(6) - 4
f(x ≤ 6) = 2x ≤ 12 - 4
f(x ≤ 6) = 2x ≤ 8
f(x ≤ 6) = x ≤ 4
We shall first of all, convert all the numbers to the same base. This means we either;
Convert the fractions to decimals OR
Convert the decimals to fractions
Let us therefore convert the decimals to fractions.
Answer:
The turns of a graph is represented by the number of maximum or minimum that the function has.
If we differenciate f(x) we get:
f'(x)=4x^3+6x
f'(x)=2x(2x^2 + 3)
Therefore f'(x) =0, when x=0. Given that negative roots are not defined.
Therefore, the number of turns will be given by the number of solutions of f'(x) which is 1.
Attached you find the graph of the function which confirms the number of turns.
If the function had other solutions, the maximum number of turns it could have is 3! because f'(x) is a third degree polynomial, therefore it can't have more than 3 solutions!
You need to simplify it first.

Use distribution

y=mx+b
m=slope
your slope is -4 :)