Questions: Tabular representations for the functions (p1 q), and (r) are given below. Write (q(z)) and (r(z)) as transformations of (p(z)), (x) -1 0 1 2 3 (p(x)) -1 0 1 8 27 (x) -1 0 1 2 3 (q(x)) 0 1 2 9 28 (x) -3.5 -2.5 -1.5 -0.5 0.5 (r(x)) -1 0 1 8 27 Selodone OA (q(x)=p(x)+1), (r(x)=p(x-2.5)) OB (q(x)=p(x)-1), (r(x)=p(x+2.5)) OC (q(x)=p(x)-1), (r(x)=p(x-2.5)) OD (q(x)=p(x)+1), (r(x)=p(x+2.5))

Solution Steps

We compare the values of \( p(x) \) and \( q(x) \) from the tables:

\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -1 & 0 & 1 & 2 & 3 \\ \hline p(x) & -1 & 0 & 1 & 8 & 27 \\ \hline q(x) & 0 & 1 & 2 & 9 & 28 \\ \hline \end{array} \]

For each \( x \), \( q(x) = p(x) + 1 \). For example:

• At \( x = -1 \), \( p(-1) = -1 \) and \( q(-1) = 0 \), so \( q(-1) = p(-1) + 1 \).
• At \( x = 0 \), \( p(0) = 0 \) and \( q(0) = 1 \), so \( q(0) = p(0) + 1 \).

Thus, \( q(x) = p(x) + 1 \).

We compare the values of \( p(x) \) and \( r(x) \) from the tables:

\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -3.5 & -2.5 & -1.5 & -0.5 & 0.5 \\ \hline r(x) & -1 & 0 & 1 & 8 & 27 \\ \hline \end{array} \]

Notice that the values of \( r(x) \) match the values of \( p(x) \) when \( x \) is shifted by \( +2.5 \). For example:

• At \( x = -3.5 \), \( r(-3.5) = -1 \), which matches \( p(-3.5 + 2.5) = p(-1) = -1 \).
• At \( x = -2.5 \), \( r(-2.5) = 0 \), which matches \( p(-2.5 + 2.5) = p(0) = 0 \).

Thus, \( r(x) = p(x + 2.5) \).

From Step 1 and Step 2, we have:

• \( q(x) = p(x) + 1 \),
• \( r(x) = p(x + 2.5) \).

Comparing these with the options:

• OA: \( q(x) = p(x) + 1 \), \( r(x) = p(x - 2.5) \) → Incorrect for \( r(x) \).
• OB: \( q(x) = p(x) - 1 \), \( r(x) = p(x + 2.5) \) → Incorrect for \( q(x) \).
• OC: \( q(x) = p(x) - 1 \), \( r(x) = p(x - 2.5) \) → Incorrect for both.
• OD: \( q(x) = p(x) + 1 \), \( r(x) = p(x + 2.5) \) → Correct for both.

Final Answer