Answer: y' = (-x)^3 - 4
Explanation:
The graph y = f(x) = x^3 passes trough the point (0,0)
because when x = 0
f(0) = y = 0^3 = 0
in the graph of the image, we can see that the graph intersects the y-axis in the point (0, -4)
This means that we have a vertical displacement of 4 units downwards.
When we have a graph y = f(x), a vertical translation of A units can be written as
y' = f(x) + A
If A is positive, the displacement is upwards, if A is negative, the displacement is downwards.
So if the displacement is of 4 units down, A = -4
We also have that when x is negative, the value of y is positive.
But in our original function, we have that for x = -1, y = (-1)^3 = -1
so in our original function, when x is negative also does y.
Then we also did a reflection around the y-axis, this means that we now evaluate the function in -x instead of x.
So the equation of the graph is:
y' = (-x)^3 - 4