The graph of the parent function f(x) = x3 is translated to form the graph of g(x) = (x + 3)3 – 4. The point (0, 0) on the graph of f(x) corresponds to which point on the graph of g(x)?

Answer 1

Let's start by seeing the difference between the original function f(x) and the transformed function g(x).

f(x) = x^3

g(x) = (x+3)^3 - 4

Both of these are cubic functions (because x is raised to the 3rd power). But, there are two differences between them.

Difference 1: x became x+3. This means that x is going to be shifted 3 units to the LEFT. If it was x-3, that would mean that x was being shifted 3 units to the RIGHT. This may seem counter-intuitive, so let's think about it. In the original equation, if y = 27, x =3. But in the (x+3) equation, if y is 27, then x+3 = 3, which means x is only 0. So x has been shifted 3 units to the LEFT. Adding 3 to x means that you need 3 more just to get to the same x spot you would be in the original for any given y value. So that's why you end up sliding backwards when you add to the x value.

Difference 2: -4 was added to the right side of the equation. This means that y is going to be shifted 4 units DOWN. Let's think about it: The TOTAL value of the right side of the equation is equal to Y. Whatever X is, or X^3, we're going to be subtracting 4 from that, every single time, when we add a -4 to the end of the equation. This drags the graph down.

In summary: The graph will be shifted 3 units left, and 4 units down.

1. If we start at (0,0), and go left, we're traveling along the x axis. If we go left from 0, that's into negative territory. We're going 3 units left, so our X coordinate will be -3.
2. If we start at (0,0) and go down, we're traveling along the y axis. If we go down from 0, we're traveling into negative territory. We're going 4 units down, so our Y coordinate will be negative 4.

When writing a point, we always put X first, and then Y. Putting it together, our point will be at (-3, -4).