Question

the function $f(x) = x^3$ is translated such that the function describing the translated graph is $g(x) = (x + 5)^3 + 2$. where is the point $(0, 0)$ for the function $f$ now located on the function $g$?
$\boldsymbol{(\square,\\ \square)}$

Explanation:

Step1: Recall function translation rules

For a function \( y = f(x) \), the transformation \( y = f(x + h) + k \) represents a horizontal shift of \( h \) units (left if \( h>0 \), right if \( h<0 \)) and a vertical shift of \( k \) units (up if \( k>0 \), down if \( k<0 \)). Here, \( f(x)=x^3 \), and \( g(x)=(x + 5)^3+2=f(x + 5)+2 \). So, the horizontal shift \( h=- 5 \) (wait, no: \( x+5=x-(-5) \), so horizontal shift is 5 units to the left, and vertical shift is 2 units up.

Step2: Apply shift to the point (0,0)

The original point is \( (x,y)=(0,0) \) on \( f(x) \). For the horizontal shift: if we have \( f(x+5) \), the \( x \)-coordinate of the point on \( g(x) \) is found by considering that \( x_{\text{new}}=x_{\text{old}}-5 \)? Wait, no. Let's think in terms of the transformation. If \( g(x)=f(x + 5)+2 \), then to get the graph of \( g(x) \) from \( f(x) \), we replace \( x \) with \( x + 5 \) (which shifts the graph left by 5 units) and then add 2 (shifts up by 2 units). So, a point \( (a,b) \) on \( f(x) \) will be transformed to \( (a - 5,b + 2) \)? Wait, no. Wait, let's take an example. For \( f(x)=x^3 \), the point \( (0,0) \) is on \( f(x) \) because \( f(0)=0^3 = 0 \). Now, for \( g(x)=(x + 5)^3+2 \), we want to find the point corresponding to \( x = 0 \) in \( f(x) \). Let's let \( t=x + 5 \), so when \( x=0 \) in \( f(x) \), we have \( f(0)=0 \). In \( g(x) \), we have \( g(x)=f(x + 5)+2 \), so we need to find \( x \) such that \( x+5 = 0 \) (to get the same input to \( f \) as when \( f(0) \))? Wait, no. Wait, the transformation is: if \( (x,y) \) is on \( f(x) \), then \( (x - h,y + k) \) is on \( f(x - h)+k \)? Wait, maybe a better way: the vertex (or the point (0,0) for \( f(x)=x^3 \)) will be shifted. The general rule for horizontal shift: if you have \( y = f(x + c) \), the graph is shifted left by \( c \) units. So, for \( f(x) \) to \( f(x + 5) \), the point (0,0) on \( f(x) \) will move to \( (-5,0) \) on \( f(x + 5) \) (because to get \( f(0) \) in \( f(x + 5) \), we need \( x+5=0\implies x=-5 \), so \( f(-5 + 5)=f(0)=0 \), so \( f(x + 5) \) has the point \( (-5,0) \)). Then, we add 2 to get \( g(x)=f(x + 5)+2 \), so we shift up by 2 units. So the \( y \)-coordinate becomes \( 0 + 2=2 \). So the point (0,0) on \( f(x) \) moves to \( (-5,2) \) on \( g(x) \). Let's verify: \( g(-5)=(-5 + 5)^3+2=0^3+2 = 2 \), which matches. So the \( x \)-coordinate is \( 0-5=-5 \) (since we shifted left by 5 units) and the \( y \)-coordinate is \( 0 + 2=2 \) (shifted up by 2 units).

Answer:

\((-5, 2)\)