Analyzing and Graphing Rational Functions

And yet despite the look on my face, you’re still talking and thinking that I care, Anonymous

Recall

Definition. A function f is a rule, relationship, or correspondence that assigns to each element x in a set D, x ∈ D (called the domain) exactly one element y in a set E, y ∈ E (called the codomain or range).

• The pair (x, y) is denoted as y = f(x) where: x is the independent variable (input) and y is the dependent variable (output). Often, both the domain D and codomain E are the set of real numbers ℝ or subsets of ℝ.

• D is the domain, the set of all possible inputs. E is the codomain or range, the set of all possible outputs.

• Key property💡: Each input has exactly one output. (No input is assigned two different outputs — this is the vertical line test!)

• Examples: constant, f(x) = c, horizontal line, slope = 0; linear, f(x) = mx + b, straight line, constant slope m and y-intercept b; quadratic $f(x) = ax^2 + bx + c$, u-shaped or inverted U, opens up (a > 0) or down (a < 0), vertex at x = $\frac{-b}{2a}$, symmetry about vertical axis through vertex; polynomial, $f(x) = a_n x^n + \dots + a_0$, a smooth and continuous curve, n roots (counting multiplicity), end behaviour determined by its leading term $a_n x^n$; exponential function, $f(x) = a \cdot b^x, a \ne 0, b \gt 0$, rapid growth (b > 1) or decay (0 < b < 1); trigonometric functions, $\sin(x), \cos(x), \tan (z)$ oscillatory, periodic behavior (period 2π for sin/cos, π for tan), sin and cos are bounded between -1 and 1, but tan is unbounded; step function $f(x) = \lfloor x \rfloor$, greatest integer ≤ x, constant on intervals [n, n+1), jumps at integers, its graph is a staircase shape; absolute value f(x) = |x|, V-shaped graph, slope changes at 0.

• Functions can be expressed in multiple forms, each useful in different contexts: verbal description, table of values (list of pairs), algebraic formula, graph, piecewise definition, recursive definition, parametric or integral form, and series representation.

• Evaluating a function means finding or computing the output value f(x) for a given input value x. f(x) = $x^2-2x +4, f(2) = 2^2 -2\cdot 2 + 4 = 4 - 4 + 4 = 4, f(0) = 0^2 -2\cdot 0 + 4 = 4, f(1) = 1^2 -2\cdot 1 + 4 = 1 -2 +4 = 3$

• The x-intercept is any point on the graph that intersects or crosses the x-axis. In other words, it is the value of x when the function (y-coordinate or y-value) is zero. The y-intercept is the point where the graph intersects or crosses the y-axis. y-coordinate of the point whose x-coordinate is 0, e.g., 2x - 3y = 6. x-intercept: set y = 0 → 2x = 6 ⇒ x=3, so (3, 0). y-intercept: set x = 0 → −3y = 6 ⇒ y = −2, so (0, −2).

• f is said to have a local or relative maximum at c if there exists an interval (a, b) containing c ($c \in (a, b)$) such that f(c) ≥ f(x) $∀ x \in (a, b) ∩ D$. f is said to have a local or relative minimum at c if there exists an interval (a, b) containing c ($c \in (a, b)$) such that f(c) ≤ f(x) $∀ x \in (a, b) ∩ D$.💡Local extrema can only occur where the function stops rising or falling —either because the derivative is zero, the derivative doesn't exist, or you're at the edge of the domain.

• First Derivative Test. Let f be differentiable on an open interval containing c, except possibly at c itself, and let c be a critical point (so f′(c) = 0 or f′ is undefined). Then:

1. If f′(x) > 0 for x just to the left of c and f′(x) < 0 for x just to the right of c, then f(c) is a local maximum.
2. If f′(x) < 0 for x just to the left of c and f′(x) > 0 for x just to the right of c, then f(c) is a local minimum.
3. If f′(x) does not change sign at c (stays positive or stays negative), then f(c) is not a local extremum.

• Second Derivative Test. If the first derivate of a function at a point c is zero (f'(c) = 0) and the second derivate at this point exists, then: (i) If f''(c) > 0, then f(c) is a local minimum. (ii) If f''(c) < 0, then f(c) is a local maximum. (iii) If f''(c) = 0, the test is inconclusive.
• Vertical asymptotes are vertical lines (perpendicular to the x-axis) of the form x = a (where a is a constant) near which the function grows without bound when x approaches a from at least one side. The line x = a is a vertical asymptote of f if at least one of the one-sided limits is infinite: $\lim_{x \to a^{-}}f(x)=\pm\infty$ or $\lim_{x \to a^{+}}f(x)=\pm\infty$.
• Horizontal asymptotes are horizontal lines (y = c, parallel to the x-axis) that the graph of the function approaches as x grows very large (positively or negatively). In other words, the function values get arbitrarily close to c as long as x is sufficiently large or more formally: $\lim_{x \to \infty}f(x)=c$ and/or $\lim_{x \to -\infty}f(x)=c$.
• A line y = m+b is an oblique (slant) asymptote as $x \to \infty$ if $\lim_{x\to\infty} [f(x) - (mx+b)] = 0.$ This means that the curve of f(x) gets more and more closer to the line y = mx + b as x grows large. For an asymptote y = mx + b as $x \to \infty, m = \lim_{x\to\infty} \frac{f(x)}{x}, b = \lim_{x\to\infty} \bigl(f(x) - m x\bigr)$, provided both limits exist.

Analysis and Plot of some functions

• $\frac{2x+1}{x^2-9}$ (Figure iv).

1. Domain: $\mathbb{R} \setminus \{ -3, 3 \}$
2. Range: All real numbers $\mathbb{R}$ (since horizontal asymptote is $y=0$ but function crosses it).
3. x-intercept: Set $2x+1=0 \leadsto x = -\frac{1}{2}$ → $\left(-\frac{1}{2}, 0\right)$. y-intercept: $f(0) = \frac{1}{-9} = -\frac{1}{9} \leadsto \left(0, -\frac{1}{9}\right)$.
4. Symmetry: None (neither even nor odd).
5. Asymptotes: $\lim_{x \to -3⁺} \frac{2x+1}{x^2-9} = \frac{-5}{0^-} = +∞, \lim_{x \to -3⁻} \frac{2x+1}{x^2-9} = \frac{-5}{0^+} = -∞$ ⇒ x = -3 is a vertical asymptote. Similarly, $\lim_{x \to 3⁺} \frac{2x+1}{x^2-9} = \frac{7}{0⁺} = +∞, \lim_{x \to 3⁻} \frac{2x+1}{x^2-9} = \frac{7}{0⁻} = -∞$ ⇒ x = 3 is a vertical asymptote, too. Since degree of numerator (1) < degree of denominator (2): $\lim_{x \to +∞} \frac{2x+1}{x^2-9} = \lim_{x \to -∞} \frac{2x+1}{x^2-9} = 0$ ⇒ y = 0 is a horizontal asymptote.

   $\lim\limits_{x \to -\infty} \frac{2x+1}{x^2-9} = 0^-$ (approaches from below). $\lim\limits_{x \to +\infty} \frac{2x+1}{x^2-9} = 0^+$ (approaches from above, $\frac{2x+1}{x^2-9} \sim \frac{2x}{x^2}=\frac{2}{x}$). For any sufficiently large 𝑥 (specifically, 𝑥 > 3), both the numerator (2x + 1) and the denominator ($x^2 -9$) are positive. Since the function is positive for large 𝑥 and the limit is 0, it approaches $0^+$.

6. Function Behavior Analysis. $f'(x) = \frac{-2(x^2+x+9)}{(x^2-9)^2}$. Numerator: The quadratic $x^2+x+9$ has a discriminant $\Delta = b^2-4ac = 1^2-4(1)(9)=-35 \lt 0$ (no real roots) and the leading coefficient (1) is positive, therefore $x^2+x+9 > 0$, and the numerator is always negative for all real x, so $f'(x) < 0$ everywhere in domain ($(x^2-9)^2 \gt 0$). The function is always decreasing on each interval.
7. Sign Analysis:
   

   Interval	$2x+1$	$x^2-9$	$f(x)$	Behavior
   $x < -3$	–	+	–	Negative
   $-3 < x < -\frac{1}{2}$	–	–	+	Positive
   $-\frac{1}{2} < x < 3$	+	–	–	Negative
   $x > 3$	+	+	+	Positive

8. Three distinct branches: (i) Left Branch ($x < -3$): Approaches y = 0 from below as $x \to -\infty$. It is always negative, decreasing, and plunges to $-\infty$ as $x \to -3^-$. (ii) Middle Branch ($-3 < x < 3$): Emerges from $+\infty$ as $x \to -3^+$, decreases continuously, crosses the x-axis at $(-\frac{1}{2}, 0)$, then crosses the y-axis at $(0, -\frac{1}{9})$, and plunges to $-\infty$ as $x \to 3^-$. (iii) Right Branch (x > 3): Emerges from $+\infty$ as $x \to 3^+$. It is always positive, always decreasing, and approaches y = 0 from above as $x \to +\infty$.  

• $\frac{x-2}{x^2-4}$ (Figure v)

1. For $x \ne \pm 2$, we have: $\frac{x-2}{x^2-4} = \frac{x-2}{(x-2)(x+2)} = \frac{1}{x+2}$
2. Domain: $\mathbb{R} \setminus \{ -2, 2 \}$.
3. Range: All real numbers except $0$ ($\mathbb{R} \setminus \{ 0 \}$).
4. x-intercept: None (numerator zero only at x = 2, which is a hole); y-intercept: $f(0) = \frac{1}{2}$ → $(0, \frac{1}{2})$
5. Asymptotes and Discontinuities. $\lim_{x \to -2⁺} \frac{x-2}{x^2-4} = \lim_{x \to -2⁺} \frac{1}{x+2} = \frac{1}{0⁺} = +∞, \lim_{x \to -2⁻} \frac{x-2}{x^2-4} = \lim_{x \to -2⁻} \frac{1}{x+2} = \frac{1}{0⁻} = -∞$ ⇒ x = -2 is a vertical asymptote. $\lim_{x \to +∞} \frac{x-2}{x^2-4} = \lim_{x \to -∞} \frac{x-2}{x^2-4} = \lim_{x \to ±∞} \frac{1}{x+2} =0$ ⇒ y = 0 is a horizontal asymptote. Besides, $\lim\limits_{x \to +\infty} f(x) = 0^+$ (approaches from above). $\lim\limits_{x \to -\infty} f(x) = 0^-$ (approaches from below).
6. $\lim\limits_{x \to 2} f(x) = \frac{1}{4}$. At x = 2 the original function is undefined because the denominator is zero, so 2 is not in the domain. The limit exists and is finite, so f has a removable discontinuity (a hole) at x = 2, located at the point (2, 1/4).
7. Sign Analysis. $x < -2: f(x) < 0$ (negative). $-2 < x < 2: f(x) > 0$ (positive). $x > 2: f(x) > 0$ (positive).
8. Derivative: $f'(x) = -\frac{1}{(x+2)^2} < 0, \forall x \in D(f)$. Then, f is strictly decreasing on each interval of its domain.
9. Second Derivative: $f''(x) = \frac{2}{(x+2)^3}$. $x < -2: f''(x) < 0$, then f concave down. $x > -2: f''(x) > 0$, then f concave up.
10. Plot: Two main branches with a gap (hole). Left Branch (x < -2): Approaches y = 0 from below as $x \to -\infty$, it is negative, strictly decreasing, and concave down, and plunges to $-\infty$ as $x \to -2^-$. Right Branch (x > -2, except x = 2): Emerges from $+\infty$ as $x \to -2^+$. It is positive, strictly decreasing, and concave up. It passes through $(0, \frac{1}{2})$ and $(1, 1/3)$, and has a hole at $(2, 1/4)$. Then, it continues for x > 2, decreasing toward $0^+$ as $x \to +\infty$ (approaches zero from above).

• $\frac{x^{2}}{1-x^{2}}$

1. Domain: ℝ - {1, -1}. f(0) = 0.
2. x-intercept. $\frac{x^{2}}{1-x^{2}} = 0 \leadsto x^2 = 0 \implies x = 0$, y-intercept, f(0) = 0/1 = 0. (0, 0) is the only intercept.
3. f(-x)=f(x), so this function is even. Symmetric about y-axis.
4. $\lim_{x \to \pm\infty}\frac{x^{2}}{1-x^{2}}=-1$, y = -1 is a horizontal asymptote.
5. $\lim_{x \to 1^{+}}\frac{x^{2}}{1-x^{2}}=\frac{1}{0^{-}} = -\infty, \lim_{x \to 1^{-}}\frac{x^{2}}{1-x^{2}}=\frac{1}{0^{+}} = \infty$. Similarly, $\lim_{x \to -1^{+}}\frac{x^{2}}{1-x^{2}}=\frac{1}{0^{+}} = \infty, \lim_{x \to -1^{-}}\frac{x^{2}}{1-x^{2}}=\frac{1}{0^{-}} = -\infty$, hence x = -1 and x = 1 are vertical asymptotes.
6. Critical points. $f'(x) = \frac{(2x)(1-x^2) - x^2(-2x)}{(1-x^2)^2} = \frac{2x - 2x^3 + 2x^3}{(1-x^2)^2} = \frac{2x}{(1-x^2)^2}$. $f'(x) = 0 \implies 2x = 0 \implies x = 0.$
7. Sign analysis. x < 0, f’(x) < 0 (decreasing), x > 0, f’(x) > 0 (increasing).
8. Local minimum: (0, 0). It is not an absolute minimum. f goes to -∞ near asymptotes.
9. Concavity analysis: $f''(x) = \frac{2(1-x^2)^2 - 2x \cdot 2(1-x^2)(-2x)}{(1-x^2)^4} = \frac{2(1-x^2) + 8x^2}{(1-x^2)^3} = \frac{2 + 6x^2}{(1-x^2)^3}$. x < -1: $(1-x^2)^3 < 0$, numerator > 0, then f’’(x) < 0 (concave down); -1 < x < 1: $(1-x^2)^3 > 0$, then f’’(x) > 0 (concave up); x > 1: $(1-x^2)^3 < 0$, then f’’(x) < 0 (concave down).
10. The graph has 4 branches (Figure 1.d.): x < -1, approaches y = -1 from below as x → -∞ and drops (it is always decreasing, concave down) to -∞ at x = -1⁻; −1 < x < 0, drops (it is always decreasing, concave up) from +∞ at x = -1⁺ to 0 at x = 0; 0 < x < 1, rises from 0 at x = 0 to +∞ at x = 1⁻ (increasing, concave up); x > 1, rises from -∞ at x = 1⁺ to -1 as x → ∞ (increasing, concave down).

    $f(x)-L=\frac{x^{2}}{1-x^{2}}-(-1)=\frac{x^{2}}{1-x^{2}}+1=\frac{x^{2}+(1-x^{2})}{1-x^{2}}=\frac{1}{1-x^{2}}$. As $x\rightarrow -\infty, x^{2}$ becomes a very large positive number, making the denominator $1-x^{2}$ a large negative number. Therefore, the fraction $\frac{1}{1-x^{2}}$ is always negative for sufficiently large negative x.

     

• $\frac{x^{3}+4}{x^{2}}$

1. Domain = ℝ - {0}. f is not defined at x = 0. Range: All real numbers.
2. x-intercept: Set numerator = 0, $x^3 + 4 = 0 \implies x^3 = -4 \implies x = -\sqrt[3]{4} \approx -1.5874$, $(-\sqrt[3]{4}, 0)$. y-intercept: None (function undefined at x = 0).
3. No symmetry (neither even nor odd).
4. $\lim_{x \to 0}\frac{x^{3}+4}{x^{2}} = \frac{4}{0^{+}} = +\infty$, x = 0 (y-axis) is a vertical asymptote.
5. Horizontal Asymptotes. None, since $\lim_{x \to \infty}\frac{x^{3}+4}{x^{2}} = \infty, \lim_{x \to -\infty}\frac{x^{3}+4}{x^{2}} = -\infty$.
6. When a linear asymptote is not parallel to the x- or y-axis, it is called an oblique asymptote. A function f(x) is asymptotic to y = mx + n (m ≠ 0) if: m = $\lim_{x \to \pm\infty} \frac{f(x)}{x} = m, \lim_{x \to \pm\infty} (f(x) -mx) = n.$ m = $\lim_{x \to \pm\infty}\frac{\frac{x^{3}+4}{x^{2}}}{x} = \lim_{x \to \pm\infty}\frac{x^{3}+4}{x^{3}}=1, lim_{x \to \pm\infty} \frac{x^{3}+4}{x^{2}} -x=lim_{x \to \pm\infty} \frac{4}{x^{2}} = 0,$ y =[m = 1, n = 0] x is an oblique asymptote.

   Since $f(x) = x + \frac{4}{x^2}$, $\forall x \ne 0, \frac{4}{x^2} > 0 \implies f(x) \gt x$. The function always lies above its oblique asymptote y = x.

7. Critical points. Rewrite: $f(x) = x + 4x^{-2} \implies f'(x) = 1 - 8x^{-3} = 1 - \frac{8}{x^3}$. $f'(x) = 0 \implies 1 - \frac{8}{x^3} = 0 \implies x^3 = 8 \implies x = 2$.
8. Monotonicity. (i) x < 0: $x^3 < 0$, so $\frac{8}{x^3} < 0 \implies f'(x) = 1 - (\text{negative}) > 0$; (ii) 0 < x < 2: $x^3 < 8$, so $\frac{8}{x^3} > 1 \implies f'(x) < 0$; (iii) x > 2: $x^3 > 8$, so $\frac{8}{x^3} < 1 \implies f'(x) > 0$. Increasing on: $(-\infty, 0)$ and $(2, \infty)$. Decreasing on (0, 2).
9. Local minimum at x = 2: $f(2) = \frac{8+4}{4} = 3$, (2, 3).
10. Concavity. $f''(x) = 24x^{-4} = \frac{24}{x^4}, f''(x) > 0, \forall x \neq 0 \text{ since } x^4 > 0$. Concave up on: $(-\infty, 0)$ and $(0, \infty)$. No inflection points (never changes concavity).
11. The plot is shown in Figure 1.c. and consists of two disconnected branches. Left branch (x < 0), starts at $-\infty$ as $x \to -\infty$ above the line y = x, crosses the x-axis at $(-\sqrt[3]{4}, 0)$, increases continuously, approaches $+\infty$ as $x \to 0^-$, and is concave up. Right branch (x > 0), starts at $+\infty$ as $x \to 0^+$, decreases to a local minimum at (2, 3), then increases to $+\infty$ as $x \to \infty$ above the line y = x. It is concave up and never touches the asymptote y = x.

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