Highway Interchange Overpass Alignment (Civil Engineering) Solution

1. Domain and Vertical Asymptote Analysis

1. Name the Governing Formulas: The Rational Function Domain Restriction Rule and Vertical Asymptote Definition: (These principles are formalized in Vertical Asymptotes.)
2. Determine Domain: A rational function is undefined where its denominator equals zero:
   $$x + 10 = 0 \implies x = -10$$
   The algebraic domain across all real numbers is:
   $$\text{Domain: } (-\infty, -10) \cup (-10, \infty)$$
3. Evaluate Vertical Asymptote: Since the numerator at $x = -10$ is non-zero:
   $$0.04(-10)^2 + 6(-10) + 80 = 0.04(100) - 60 + 80 = 4 - 60 + 80 = 24 \neq 0$$
   The line $x = -10$ is a vertical asymptote.
4. Physical Highway Context: Because the connector ramp begins at station $x = 0$ and extends forward into positive distance ($x \ge 0$), the singularity at $x = -10$ (which is $1,000\text{ m}$ behind the start point) lies completely outside the physical operational highway domain. The elevation curve is continuous and smooth for all physical stationing $x \ge 0$.

2. Elevation at Start Station ($y$-Intercept)

Evaluate the elevation function at station $x = 0$:

3. Polynomial Long Division & Slant Asymptote Extraction

1. Name the Governing Formula: The Slant Asymptote Decomposition Formula via Polynomial Division:
   $$f(x) = (mx + b) + \frac{R(x)}{Q(x)}$$
   (This procedure is defined in Slant Asymptotes and Polynomial Division.)
2. Perform Long Division: Divide $(0.04x^2 + 6x + 80)$ by $(x + 10)$:
   • Leading term quotient: $\frac{0.04x^2}{x} = 0.04x$.
   • Multiply and subtract:
     $$(0.04x^2 + 6x + 80) - 0.04x(x + 10) = (0.04x^2 + 6x + 80) - (0.04x^2 + 0.4x) = 5.6x + 80$$
   • Next quotient term: $\frac{5.6x}{x} = 5.6$.
   • Multiply and subtract:
     $$(5.6x + 80) - 5.6(x + 10) = (5.6x + 80) - (5.6x + 56) = 24$$
3. Write Quotient-Remainder Form:
   $$y(x) = (0.04x + 5.6) + \frac{24}{x + 10}$$
4. Extract Slant Asymptote: As station distance $x \to \infty$:
   $$\lim_{x \to \infty} \frac{24}{x + 10} = 0$$
   Therefore, the slant (oblique) asymptote is the linear equation:
   $$y_{\text{asymptote}} = 0.04x + 5.6$$

4. Engineering Interpretation

• Slope ($m = 0.04$): Because $x$ is measured in hundreds of meters ($1\text{ unit} = 100\text{ m}$) and $y$ is in meters, a slope of $0.04$ represents a constant long-range grade of:
  $$\text{Grade} = \frac{0.04\text{ m elevation}}{100\text{ m horizontal}} \times 100\% = 0.04\% \quad (\text{or } 4\text{ cm vertical per } 100\text{ m})$$
  This establishes the steady drainage and climb gradient for the mainline highway connection.
• Intercept ($b = 5.6\text{ m}$): Represents the projected baseline datum elevation of the continuous highway corridor if the constant grade were extrapolated back to station $x = 0$.

5. Deck Elevation Approach Direction

Examine the remainder term for positive stations $x \ge 0$: