Question

Analyze the effect of the leading coefficient on j(x)=2513x−−−√+4
in comparison to k(x)=13x−−−√+4
. Enter your response in fraction form if needed.(1 point)

Answer 1

To analyze the effect of the leading coefficient on the functions \( j(x) = \sqrt{2513x} + 4 \) and \( k(x) = \sqrt{13x} + 4 \), we will compare their leading coefficients and how they affect the growth rates of each function as \( x \) increases.

1. **Leading Coefficients**:
- In \( j(x) \), the leading coefficient is \( \sqrt{2513} \).
- In \( k(x) \), the leading coefficient is \( \sqrt{13} \).

2. **Comparing Growth Rates**:
- The dominant term in both functions as \( x \) becomes very large is the square root term (i.e., \( \sqrt{2513x} \) and \( \sqrt{13x} \)).
- We can express the functions as:
\[
j(x) \approx \sqrt{2513} \cdot \sqrt{x} + 4
\]
\[
k(x) \approx \sqrt{13} \cdot \sqrt{x} + 4
\]
- The term \( +4 \) becomes negligible compared to the square root as \( x \) increases.

3. **Ratio of Leading Coefficients**:
- To quantify the effect, we can look at the ratio of the leading coefficients:
\[
\frac{j(x)}{k(x)} \approx \frac{\sqrt{2513} \cdot \sqrt{x}}{\sqrt{13} \cdot \sqrt{x}} = \frac{\sqrt{2513}}{\sqrt{13}}
\]
- This ratio remains constant as \( x \) approaches infinity and gives us a clear idea of how much faster \( j(x) \) grows compared to \( k(x) \).

4. **Simplifying the Fraction**:
- We can simplify the ratio further:
\[
\frac{\sqrt{2513}}{\sqrt{13}} = \sqrt{\frac{2513}{13}}
\]
- We can compute \( \frac{2513}{13} \):
\[
2513 \div 13 \approx 193.3076923
\]

Thus, the key effect of the leading coefficient in this context highlights that \( j(x) \) will grow significantly faster than \( k(x) \) as \( x \) increases, and specifically:

**The effect of the leading coefficient can be expressed as**:
\[
\sqrt{\frac{2513}{13}}
\]

This tells us that \( j(x) \) increases approximately \( \sqrt{\frac{2513}{13}} \) times faster than \( k(x) \), which numerically evaluates to around \( 13.155 \), making \( j(x) \) significantly steeper than \( k(x) \) for large \( x \).