Graphing Rational Functions Worksheet 2

Let’s solve each problem step by step. We’ll find vertical asymptotes (VA), horizontal asymptotes (HA), domain, holes, intercepts, and continuity for each rational function.

---

Problem 1:
Function:
\[
f(x) = \frac{x^2 + 4x - 5}{x^2 + 9x + 20}
\]

Step 1: Factor numerator and denominator.

Numerator: \(x^2 + 4x - 5 = (x + 5)(x - 1)\)

Denominator: \(x^2 + 9x + 20 = (x + 5)(x + 4)\)

So,
\[
f(x) = \frac{(x + 5)(x - 1)}{(x + 5)(x + 4)}
\]

Cancel common factor \((x + 5)\), but note: this creates a hole at \(x = -5\) (since it makes original denominator zero).

Simplified function:
\[
f(x) = \frac{x - 1}{x + 4}, \quad x \ne -5
\]

→ Vertical Asymptote (VA): Set simplified denominator = 0 → \(x + 4 = 0\) → \(x = -4\)

→ Horizontal Asymptote (HA): Degrees of numerator and denominator are both 1 → ratio of leading coefficients: \(1/1 = 1\) → \(y = 1\)

✔ VA: \(x = -4\)
✔ HA: \(y = 1\)

*(Note: Hole at x = -5 is not asked here — only VA and HA)*

---

Problem 2:
Function:
\[
f(x) = \frac{x^2 - 9}{x + 3}
\]

Factor numerator: \(x^2 - 9 = (x - 3)(x + 3)\)

So,
\[
f(x) = \frac{(x - 3)(x + 3)}{x + 3}
\]

Cancel \((x + 3)\), but hole at \(x = -3\)

Simplified: \(f(x) = x - 3\), with hole at \(x = -3\)

→ This is now a linear function (degree 1 over degree 0 after simplifying? Wait — actually, after canceling, it's just a line — so no vertical asymptote!

But let’s think carefully:

Original denominator: \(x + 3 = 0\) → \(x = -3\)

After canceling, we have no denominator left → so no vertical asymptote, only a hole.

Horizontal asymptote: After simplification, it’s a line → as x → , y → ∞ → no horizontal asymptote

Wait — that’s correct? Let me double-check.

Actually, for rational functions, if degree of numerator > degree of denominator after simplifying, there is NO horizontal asymptote (there might be an oblique/slant asymptote, but the question only asks for HA).

In this case, after simplifying, it’s a polynomial (linear), so:

→ VA: None (because the factor canceled out — only a hole)

→ HA: None (since it behaves like a line going to infinity)

But wait — sometimes teachers expect you to look at original degrees before simplifying? No — standard rule: simplify first, then analyze.

However, let’s check what happens near x = -3: original function undefined, but limit exists → hole, not asymptote.

So:

✔ VA: None
✔ HA: None

*(Some might say “DNE” or write “none”)*

---

Problem 3:
Function:
\[
f(x) = \frac{x + 6}{2x^2 + 9x - 18}
\]

First, factor denominator.

Find two numbers that multiply to \(2 * (-18) = -36\), add to 9 → 12 and -3.

So:
\[
2x^2 + 9x - 18 = 2x^2 + 12x - 3x - 18 = 2x(x + 6) -3(x + 6) = (2x - 3)(x + 6)
\]

So,
\[
f(x) = \frac{x + 6}{(2x - 3)(x + 6)}
\]

Cancel \((x + 6)\), but hole at \(x = -6\)

Simplified:
\[
f(x) = \frac{1}{2x - 3}, \quad x \ne -6
\]

→ VA: Set simplified denominator = 0 → \(2x - 3 = 0\) → \(x = \frac{3}{2}\)

→ HA: Degree of numerator (0) < degree of denominator (1) → HA is \(y = 0\)

✔ VA: \(x = \frac{3}{2}\)
✔ HA: \(y = 0\)

---

Now Problems 4 and 5 require graphing and filling blanks. Since we can’t draw graphs here, I’ll compute all required values.

---

Problem 4:
Function:
\[
y = \frac{3}{x + 2}
\]

This is already simplified.

→ Domain: All real except where denominator = 0 → \(x + 2 = 0\) → \(x = -2\) → Domain: \(x \ne -2\)

→ VA: \(x = -2\)

→ Holes: None (no common factors to cancel)

→ x-int: Set y = 0 → \(\frac{3}{x+2} = 0\) → never true → no x-intercept

→ y-int: Set x = 0 → \(y = \frac{3}{0 + 2} = \frac{3}{2}\) → (0, 1.5)

→ HA: Degree num (0) < den (1) → \(y = 0\)

→ Continuous/Discontinuous: Discontinuous at x = -2 (vertical asymptote)

✔ Domain: \(x \ne -2\)
✔ VA: \(x = -2\)
✔ Holes: None
✔ x-int: None
✔ y-int: \(\frac{3}{2}\)
✔ HA: \(y = 0\)
✔ Continuous/Discontinuous: Discontinuous

---

Problem 5:
Function:
\[
y = \frac{x^2 - 9}{x - 3}
\]

Factor numerator: \(x^2 - 9 = (x - 3)(x + 3)\)

So,
\[
y = \frac{(x - 3)(x + 3)}{x - 3}
\]

Cancel \((x - 3)\), but hole at \(x = 3\)

Simplified: \(y = x + 3\), with hole at \(x = 3\)

→ Domain: All real except \(x = 3\)

→ VA: None (factor canceled — only hole)

→ Holes: At \(x = 3\) → plug into simplified: \(y = 3 + 3 = 6\) → hole at (3, 6)

→ x-int: Set y = 0 in simplified: \(x + 3 = 0\) → \(x = -3\) → (-3, 0)

→ y-int: Set x = 0 → \(y = 0 + 3 = 3\) → (0, 3)

→ HA: Simplified function is linear → no horizontal asymptote (goes to ±∞)

→ Continuous/Discontinuous: Discontinuous at x = 3 (hole)

✔ Domain: \(x \ne 3\)
✔ VA: None
✔ Holes: (3, 6)
✔ x-int: (-3, 0)
✔ y-int: (0, 3)
✔ HA: None
✔ Continuous/Discontinuous: Discontinuous

---

Final Answer:

1. VA: \(x = -4\), HA: \(y = 1\)
2. VA: None, HA: None
3. VA: \(x = \frac{3}{2}\), HA: \(y = 0\)
4. Domain: \(x \ne -2\), VA: \(x = -2\), Holes: None, x-int: None, y-int: \(\frac{3}{2}\), HA: \(y = 0\), Continuous/Discontinuous: Discontinuous
5. Domain: \(x \ne 3\), VA: None, Holes: (3, 6), x-int: (-3, 0), y-int: (0, 3), HA: None, Continuous/Discontinuous: Discontinuous