Graphs of Common Functions
This topic covers the fundamental skill of sketching and recognising common mathematical functions without a calculator. A strong visual understanding of these graphs is essential for solving a wide range of problems, including finding roots and solving inequalities.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The overall shape of a polynomial `y = a*xn +⋯` is determined by its degree `n` (even/odd) and the sign of the leading coefficient `a`. This dictates the graph's 'end behaviour' as x approaches positive or negative infinity.
- The graph of `y = |f(x)|` is created by sketching `y = f(x)` and then reflecting any part of the graph that is below the x-axis (`y < 0`) in the x-axis.
- Exponential functions, `y = a^x` (for `a > 1`), grow rapidly and have a horizontal asymptote at `y = 0`. Their inverses, logarithmic functions `y = loga(x)`, grow slowly and have a vertical asymptote at `x = 0`.
- Be familiar with the domains of functions. For example, `y = √(x)` is only defined for `x ≥ 0`, and `y = ln(x)` is only defined for `x > 0`.
- Know the shapes and periods of trigonometric functions. `sin(x)` and `cos(x)` have a period of 2π, while `tan(x)` has a period of π and vertical asymptotes where `cos(x) = 0`.
Formulae
y = a xn (for large |x|) To determine the end behaviour of a polynomial. The term with the highest power of x dominates the shape of the graph far from the origin.
y = |f(x)| = { f(x) if f(x)≥0; -f(x) if f(x)<0 } When sketching a modulus function or solving an equation involving one. This definition splits the problem into two separate cases based on the sign of f(x).
Definitions
- Modulus Function
- The function `y = |x|` returns the non-negative value of x. When applied to a function as `y = |f(x)|`, it makes all negative outputs of `f(x)` positive.
- Asymptote
- A straight line that a curve approaches arbitrarily closely as it heads towards infinity.
- End Behaviour
- The behaviour of the graph of a function as x approaches positive infinity or negative infinity.
Worked example
By sketching the relevant graphs, or otherwise, find the number of distinct real solutions to the equation `|x2 - 1| = 2 - x`.
- 1
The equation involves a modulus, so we can solve it by finding the intersections of the graphs `y = |x2 - 1|` and `y = 2 - x`.
- 2
First, sketch `y = x2 - 1`.
This is a standard parabola shifted down by 1, with roots at x = -1 and x = 1, and a vertex at (0, -1).
- 3
Next, sketch `y = |x2 - 1|`.
This involves reflecting the part of the parabola below the x-axis (between x = -1 and x = 1) in the x-axis.
The vertex at (0, -1) becomes a local maximum at (0, 1).
The graph has a 'W' shape.
- 4
On the same axes, sketch the straight line `y = 2 - x`.
It has a y-intercept at (0, 2) and an x-intercept at (2, 0).
- 5
Visually count the intersections.
The line starts above the 'W' shape at the y-axis (y=2 vs y=1) and has a negative slope.
It will cross the right-hand arm of the 'W' once for x > 1.
It will also cross the central, inverted part of the 'W' between x = -1 and x = 1.
Let's see if it crosses the left-hand arm for x < -1.
- 6
To be certain, we solve algebraically.
Case 1:
`x2 - 1 = 2 - x` (for `x≤-1` or `x≥1`)This gives `x2 + x - 3 = 0`.
The solutions are `x = (-1 ± √(13))/2`.
`(-1+√(13))/2` is approx 1.3, which is `>1`, so this is a valid solution.
`(-1-√(13))/2` is approx -2.3, which is `< -1`, so this is also a valid solution.
- 7
Case 2:
`-(x2 - 1) = 2 - x` (for `-1<x<1`)This gives `-x2 + 1 = 2 - x`, so `x2 - x + 1 = 0`.
The discriminant is `b2 - 4ac = (-1)2 - 4(1)(1) = -3`.
Since the discriminant is negative, there are no real solutions in this case.
- 8
Combining the valid solutions from both cases, we have two distinct real solutions.
Answer: 2
Common mistakes
- ×Mistaking the sketch of `y = |f(x)|` (reflects negative outputs in x-axis) for `y = f(|x|)` (discards graph for x<0, reflects graph for x>0 in y-axis).
- ×Incorrectly identifying the end behaviour of polynomials, especially when the leading coefficient is negative. For `y = -x3`, the graph should go from top-left to bottom-right.
- ×When solving modulus equations algebraically, forgetting to check whether the solutions found are valid within the domain for that specific case (e.g., checking if `f(x)` was indeed negative for the `y = -f(x)` case).
No-calculator tips
- ✓To sketch graphs quickly, focus on key features: roots (where y=0), y-intercepts (where x=0), turning points, and asymptotes. A rough sketch is often all that is needed.
- ✓When comparing two functions, `f(x)` and `g(x)`, test simple integer values (e.g., x=0, x=1, x=-1) to quickly determine which function has a greater value in a particular region. This helps confirm your sketch.
- ✓For polynomials, the sign of the highest power's coefficient and whether the power is even or odd immediately tells you the graph's shape as it leaves the page (e.g., positive coefficient and even power means it goes to +infinity at both ends).