Suppose $f(x)=\frac{x^{2}+5 x+7}{x-7}$. Notice that $f(5)=-28.5$. What does this tell us about the numerator and denominator of $f$ ? When $x=5$, the value of $x^{2}+5 x+7$ is times as large as the value of Preview

Step 1 ：Suppose we have the function \(f(x)=\frac{x^{2}+5 x+7}{x-7}\). We are given that \(f(5)=-28.5\).

Step 2 ：This tells us that when we substitute \(x=5\) into the function, the result is \(-28.5\). Therefore, the ratio of the numerator and the denominator when \(x=5\) is \(-28.5\).

Step 3 ：Substituting \(x=5\) into the numerator, we get \(5^{2}+5*5+7=57\).

Step 4 ：Substituting \(x=5\) into the denominator, we get \(5-7=-2\).

Step 5 ：Therefore, the ratio of the numerator and the denominator when \(x=5\) is \(\frac{57}{-2}=-28.5\).

Step 6 ：This confirms that when \(x=5\), the ratio of the numerator and the denominator of the function \(f(x)\) is indeed \(-28.5\). This is consistent with the given information that \(f(5)=-28.5\).

Step 7 ：Therefore, the numerator is \(-28.5\) times as large as the denominator when \(x=5\).

Step 8 ：Final Answer: When \(x=5\), the value of \(x^{2}+5 x+7\) is \(-28.5\) times as large as the value of \(x-7\). Therefore, the final answer is \(\boxed{-28.5}\).