Question

tia states that the graph of $g(x) = (x - 2)^3 + 7$ is a translation of 2 units to the left and 7 units up from $f(x) = x^3$. is tias description of the translations correct? explain.

Explanation:

Step1: Recall translation rules

For a function \( y = f(x) \), the transformation \( y = f(x - h) + k \) means:

• If \( h>0 \), the graph shifts \( h \) units to the right; if \( h < 0 \), it shifts \( |h| \) units to the left.
• If \( k>0 \), the graph shifts \( k \) units up; if \( k < 0 \), it shifts \( |k| \) units down.

Step2: Analyze \( g(x)=(x - 2)^{3}+7 \)

Given \( f(x)=x^{3} \), and \( g(x)=(x - 2)^{3}+7 \). Comparing with \( f(x - h)+k \):

• Here, \( h = 2 \) (since \( x-2=x - h\) implies \( h = 2 \)) and \( k = 7 \).
• From the translation rule, since \( h = 2>0 \), the graph of \( f(x) \) shifts 2 units to the right (not left) when we go from \( f(x) \) to \( f(x - 2)=(x - 2)^{3} \).
• Then, adding 7 ( \( k = 7>0 \)) shifts the graph 7 units up.

Tia's description is incorrect. For a function \( y = f(x) \), the transformation \( y=f(x - h)+k \) has \( h = 2 \) (so shift 2 units right, not left) and \( k = 7 \) (shift 7 units up). The horizontal shift for \( g(x)=(x - 2)^{3}+7 \) from \( f(x)=x^{3} \) is 2 units to the right (because \( h = 2>0 \) in \( f(x - h) \)), not left, while the vertical shift (7 units up) is correct.

Answer:

Tia's description is incorrect. The graph of \( g(x)=(x - 2)^{3}+7 \) is a translation of 2 units to the right (not left) and 7 units up from \( f(x)=x^{3} \). For a function \( y = f(x) \), the transformation \( y = f(x - h)+k \) shifts the graph \( h \) units right (when \( h>0 \)) and \( k \) units up (when \( k>0 \)). Here, \( h = 2 \) (so right shift) and \( k = 7 \) (up shift).