Question

in the table, ( y_1 ) is defined by a rational function of the form ( \frac{x - p}{x - q} ). use the table to find the values of ( p ) and ( q ). ( p = square ) \\(\

$$\begin{array}{|c|c|} \\hline x & y_1 \\\\ \\hline 0 & 0.66666 \\\\ \\hline 1 & 0.5 \\\\ \\hline 2 & 0 \\\\ \\hline 3 & \\text{error} \\\\ \\hline 4 & 2 \\\\ \\hline 5 & 1.5 \\\\ \\hline 6 & 1.3333 \\\\ \\hline \\end{array}$$

\\)

Explanation:

Step1: Find q from the ERROR value

When \( x = 3 \), \( Y_1 \) has an ERROR, which means the denominator \( x - q = 0 \) at \( x = 3 \). So \( 3 - q = 0 \), which gives \( q = 3 \).

Step2: Find p using \( x = 2 \)

When \( x = 2 \), \( Y_1 = 0 \). Substitute into the function \( \frac{x - p}{x - q} \), we have \( \frac{2 - p}{2 - 3} = 0 \). The numerator must be 0 for the fraction to be 0, so \( 2 - p = 0 \), which gives \( p = 2 \).

Answer:

\( p = 2 \)