After the parabola and the straight line, the graphs that appear most often in exams are the hyperbola and the root and absolute-value functions. Questions usually concern the domain, asymptotes, intersection points of graphs and the number of solutions of equations with a parameter. Once you know the "skeleton" of each function, you can sketch its shifted versions at a glance.
§1. The hyperbola and linear-fractional functions
y = k/x, k ≠ 0 — inverse proportionality; D = E = (−∞; 0) ∪ (0; +∞)
k > 0 — branches in quadrants I and III, decreasing on each interval.
k < 0 — branches in quadrants II and IV, increasing on each interval.
The coordinate axes are asymptotes: the graph approaches them endlessly but never crosses them.
The function is odd: the graph is symmetric about the origin.
A subtle point: for k > 0, y = k/x decreases "on each interval", but not on the whole domain: −1 < 1, yet f(−1) = −k < f(1) = k. Shifted hyperbola:
y = k/(x − a) + b — asymptotes x = a and y = b
y = (ax + b)/(cx + d) — asymptotes x = −d/c and y = a/c
Worked example
Find the asymptotes of y = (2x + 1)/(x − 1) and describe the graph.
Separate the whole part: 2x + 1 = 2(x − 1) + 3, so y = 2 + 3/(x − 1).
This is y = 3/x shifted 1 unit right and 2 units up.
Asymptotes: x = 1 and y = 2; k = 3 > 0 — the branches lie in the "quadrants I and III" formed by the asymptotes.
§2. The square root function
y = √x: D = [0; +∞), E = [0; +∞), increasing
Its graph is the right half of the parabola y = x² reflected in y = x (a parabola "lying on its side"), because for x ≥ 0 the functions √x and x² are inverses.
y = √(x − a) + b — the graph "starts" at the point (a; b).
y = ∛x — defined for all real numbers, odd and increasing.
Worked example
Find D, E and the Ox intercept of y = √(x + 4) − 1.
x + 4 ≥ 0 → D = [−4; +∞).
√(x + 4) ≥ 0 → y ≥ −1, E = [−1; +∞); the graph starts at (−4; −1).
√(x + 4) = 1 → x + 4 = 1 → x = −3. Point (−3; 0).
§3. The absolute value function
|x| = x if x ≥ 0; |x| = −x if x < 0
y = |x| is a "V" with vertex (0; 0), an even function. The vertex of y = |x − a| + b is (a; b). Two key construction rules:
y = |f(x)|: the part of the graph of f below Ox is reflected upwards.
y = f(|x|): the part for x ≥ 0 is kept and mirrored to the left in Oy.
Worked example
Find the area of the triangle formed by y = |x − 2| − 3 and the Ox axis.
Vertex (2; −3), arms pointing up.
With Ox: |x − 2| = 3 → x = 5 or x = −1.
Base 6, height 3: S = 6·3/2 = 9.
Worked example (parameter)
How many solutions does |x² − 4| = a have, depending on a?
The part of y = x² − 4 below Ox (−2 < x < 2) is reflected upwards; the new "hump" has its top at (0; 4).
Intersections with the horizontal line y = a: a < 0 — 0; a = 0 — 2; 0 < a < 4 — 4; a = 4 — 3; a > 4 — 2.
§4. Power functions
y = xⁿ: n even — even function, y ≥ 0; n odd — odd function, E = ℝ
All graphs y = xⁿ (n > 0) pass through (1; 1).
For 0 < x < 1 a larger exponent gives a smaller value (x³ < x²); for x > 1 it is the other way round.
y = x⁻ⁿ = 1/xⁿ looks like a hyperbola; y = 1/x² is even, y > 0.
§5. Intersections of graphs
The common points of two graphs are the solutions of f(x) = g(x). A sketch shows the number of solutions in advance.
Worked example
How many common points do f(x) = 2√x and g(x) = 2x have, and what is the area between them?
2√x = 2x → √x = x → x = x² → x = 0 or x = 1 — two points: (0; 0) and (1; 2).
You rarely need to plot a new function from scratch: usually its graph is obtained from a familiar one (y = √x, y = |x|, y = 1/x, y = x²) by a few simple transformations. The method works in the same way for every function in the previous sections.
y = f(x − a) + b — the graph moves a units to the right and b units up
y = −f(x) — reflection in Ox; y = f(−x) — reflection in Oy
y = k·f(x) (k > 0) — stretch along Oy by factor k (a compression if k < 1)
A change inside the brackets acts horizontally and ‘the opposite way’: the graph of f(x − 3) moves right, not left, because the old point x = 0 is now reached at x = 3.
A term outside the brackets acts vertically and ‘the obvious way’: + b up, − b down.
The order matters: a shift and a reflection done in different orders can give different graphs. A safe order is: first the changes inside the brackets, then −f or k·f, and the vertical shift last.
A shift moves the domain and the range together with the graph. For example, for y = √(x + 4) − 1 we have D = [−4; +∞) and E = [−1; +∞); the graph meets Ox where √(x + 4) = 1, i.e. x = −3, and meets Oy at y = √4 − 1 = 1.
Worked example
Sketch y = 2 − |x + 1| and find the area of the triangle it forms with the x-axis.
Start from y = |x| (vertex (0; 0), arms pointing up).
|x + 1| — shift 1 unit left, vertex (−1; 0).
−|x + 1| — reflect in Ox, the arms now point down.
2 − |x + 1| — shift 2 units up, vertex (−1; 2).
Intersections with Ox: |x + 1| = 2 → x = 1 or x = −3; base 4, height 2.
S = 4·2/2 = 4.
Check yourself
Find the asymptotes of y = 3/(x + 2) − 1.
Which quadrants does y = −6/x occupy, and what is its value at x = −2?
Find the domain and range of y = √(5 − x) + 2.
Solve |x − 3| = 5.
Find the intersection points of y = 4/x and y = x.
How many solutions does |x² − 4| = 3 have?
Which shifts turn the graph of y = √x into the graph of y = √(x − 3) + 2?
Find the area of the triangle bounded by y = 1 − |x − 2| and the x-axis.
Common mistakes
Thinking the graph of y = k/x crosses the axes — the axes are asymptotes.
Calling y = 1/x decreasing on its whole domain (only on each interval separately).
Treating √x as defined for negative x; confusing it with ∛x.
Swapping the rules for |f(x)| and f(|x|).
Writing only one solution (x = 8) of |x − 3| = 5.
Shifting the graph of y = f(x − a) to the left: − a inside the brackets moves it to the right.