Rational graphs

Vertical asymptote: condition

For a rational function f(x) = P(x)/Q(x) [numerator over denominator], x = a is a vertical asymptote when the denominator Q(a) = 0 AND the numerator P(a) ≠ 0. Cancel any common factors first before testing.

Horizontal asymptote: deg(P) < deg(Q)

For f(x) = P(x)/Q(x) [numerator over denominator], if the degree (highest power) of the numerator is LESS than the degree of the denominator → horizontal asymptote is y = 0 (the x-axis)

Horizontal asymptote: deg(P) = deg(Q)

For f(x) = P(x)/Q(x) [numerator over denominator], if the degree of the numerator EQUALS the degree of the denominator → horizontal asymptote is y = a/b where a and b are the leading coefficients of the numerator and denominator respectively

Horizontal asymptote: deg(P) > deg(Q)

For f(x) = P(x)/Q(x) [numerator over denominator], if the degree of the numerator is GREATER than the degree of the denominator → no horizontal asymptote exists

Oblique/slant asymptote: condition

For f(x) = P(x)/Q(x) [numerator over denominator], an oblique asymptote exists when the degree of the numerator is exactly ONE MORE than the degree of the denominator

Oblique/slant asymptote: how to find

Perform polynomial long division on P(x) ÷ Q(x) [numerator ÷ denominator] to get: mx + b + R(x)/Q(x), where R(x) is the remainder. The oblique asymptote is the line y = mx + b — the remainder term R(x)/Q(x) vanishes as x → ±∞

Can a rational function have both a horizontal AND oblique asymptote?

No — for f(x) = P(x)/Q(x) [numerator over denominator], it can only ever have one or the other, never both simultaneously

Can a rational function have two horizontal asymptotes?

No — a rational function f(x) = P(x)/Q(x) [numerator over denominator] has at most one horizontal asymptote

Crossing a horizontal asymptote y = L: how to test

Solve f(x) = L directly — if a real solution exists, the graph crosses the horizontal asymptote at that x value. A graph CAN cross its horizontal asymptote.

Crossing an oblique asymptote y = mx + b: how to test

After long division, f(x) = mx + b + R(x)/Q(x) [where R(x) is the remainder, Q(x) is the denominator]. Solve R(x) = 0 — any solutions are where the graph crosses the oblique asymptote

y-intercept: how to find

Substitute x = 0 into f(x) and evaluate — the y-intercept is f(0), provided the denominator ≠ 0 at x = 0

x-intercept: how to find

For f(x) = P(x)/Q(x) [numerator over denominator], set the numerator P(x) = 0 and solve, but only accept solutions where the denominator Q(x) ≠ 0 at those values

Sign chart for f(x): what does f(x) > 0 mean?

The graph of f is ABOVE the x-axis in that interval

Sign chart for f(x): what does f(x) < 0 mean?

The graph of f is BELOW the x-axis in that interval

Sign chart: how to construct

1) Mark all x-intercepts [where numerator = 0] and vertical asymptotes [where denominator = 0] on a number line 2) Test the sign of f(x) by substituting any value from each interval between those marked points

Stationary points: how to find

Differentiate f(x) to get f'(x), then solve f'(x) = 0. The x values found are the stationary points (local maxima, minima, or points of inflection)

Local minimum: second derivative test

Substitute the stationary point x = a into f''(x) [the second derivative]. If f''(a) > 0, the graph is concave up at that point → local minimum

Local maximum: second derivative test

Substitute the stationary point x = a into f''(x) [the second derivative]. If f''(a) < 0, the graph is concave down at that point → local maximum

Point of inflection: full condition

Find where f''(x) = 0 [second derivative equals zero], then verify that f''(x) changes sign (+ to − or − to +) on either side of that point. Both conditions must hold — f''(x) = 0 alone is not enough.

Symmetry about the y-axis: condition

Test if f(−x) = f(x). If true, the function is EVEN and symmetric about the y-axis. Graphically: left and right sides are mirror images.

Symmetry about the origin: condition

Test if f(−x) = −f(x). If true, the function is ODD and has rotational symmetry about the origin (180° rotation looks identical).

Steps to sketch a polynomial function

1) Stationary points: solve f'(x) = 0 2) Classify each: use f''(x) > 0 (min) or f''(x) < 0 (max) 3) Concavity/inflection: find sign changes of f''(x) 4) y-intercept: evaluate f(0) 5) Sketch 6) x-intercepts only if asked: solve f(x) = 0 using Newton-Raphson

Steps to sketch a rational function f(x) = P(x)/Q(x)

1) State domain (exclude where denominator = 0) 2) Stationary points: solve f'(x) = 0 3) Classify and test concavity using f''(x) 4) y-intercept: f(0) 5) x-intercepts: set numerator = 0 6) All asymptotes: vertical (denominator = 0), horizontal or oblique (compare degrees) 7) Test symmetry: even or odd? 8) Sign chart for f(x)

Removable discontinuity (hole) in a rational function: when does it occur?

When a factor cancels from BOTH the numerator P(x) and denominator Q(x) before simplifying. The function is undefined at that x value but has no asymptote there — it appears as a hole on the graph instead.