Polynomial Roots and Intercepts
This topic covers finding where a function's graph crosses the coordinate axes, a key step in sketching. It also involves understanding the maximum and minimum possible number of x-intercepts (real roots) a polynomial can have based on its degree.
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
- To find the y-intercept of a graph y = f(x), calculate the value of y when x = 0. This gives the coordinate (0, f(0)).
- To find the x-intercepts (also called roots) of a graph y = f(x), set y = 0 and solve the equation f(x) = 0 for all possible values of x.
- A polynomial of degree 'n' can have at most 'n' distinct real roots (x-intercepts).
- A polynomial of odd degree (like a cubic) must cross the x-axis at least once, so it always has at least one real root.
- A polynomial of even degree (like a quadratic or quartic) might not cross the x-axis at all, meaning it can have zero real roots.
- If a polynomial has a repeated factor, such as (x-a)2, its graph touches the x-axis at x=a but does not cross it. This is a repeated root.
Diagram
Formulae
yintercept = f(0) To find the y-coordinate where the graph of y = f(x) crosses the y-axis.
f(x) = 0 To find the x-coordinates (roots) where the graph of y = f(x) crosses or touches the x-axis.
Definitions
- Intercept
- A point where the graph of a function crosses or touches one of the coordinate axes.
- Root
- A solution to the equation f(x) = 0. Graphically, this corresponds to an x-intercept.
- Degree of a Polynomial
- The highest power of the variable in the polynomial expression. For example, the degree of f(x) = 3x4 - x2 + 5 is 4.
Worked example
Consider the polynomial function f(x) = x3 + 2x2 - x - 2. Determine the coordinates of all points where its graph intersects the x and y axes. What is the maximum number of real roots a polynomial of this type could have?
- 1
Step 1:
Find the y-intercept.
Set x = 0 in the function:f(0) = (0)3 + 2(0)2 - (0) - 2 = -2The y-intercept is at the coordinate (0, -2).
- 2
Step 2:
Find the x-intercepts.
Set f(x) = 0:x3 + 2x2 - x - 2 = 0.
- 3
Step 3:
Factorise the polynomial.
We can factor by grouping:
x2(x + 2) - 1(x + 2) = 0.
- 4
Step 4:
Continue factorising:
(x2 - 1)(x + 2) = 0.
The first bracket is a difference of two squares.
- 5
Step 5:
Complete the factorisation:
(x - 1)(x + 1)(x + 2) = 0.
- 6
Step 6:
Solve for the roots.
The solutions are x = 1, x = -1, and x = -2.
The x-intercepts are at the coordinates (1, 0), (-1, 0), and (-2, 0).
- 7
Step 7:
Address the second part of the question.
The function is a cubic (degree 3).
A polynomial of degree 'n' has at most 'n' real roots.
Therefore, a cubic polynomial can have a maximum of 3 real roots.
Answer: The intercepts are (0, -2), (1, 0), (-1, 0), and (-2, 0). The maximum number of real roots is 3.
Common mistakes
- ×Confusing the procedures for finding x and y intercepts; for example, setting x=0 to find the roots.
- ×Stating an intercept as a single number (e.g., 'the y-intercept is 5') instead of a coordinate pair (e.g., '(0, 5)').
- ×Assuming a polynomial of degree 'n' must have exactly 'n' real roots. It can have fewer (e.g., y = x2 + 4 has degree 2 but no real roots).
- ×Struggling with the algebraic factorisation of cubic or higher-order polynomials to find the roots.
No-calculator tips
- ✓The y-intercept of any polynomial is always its constant term.
- ✓To find roots of a polynomial with integer coefficients, test factors of the constant term. This is a quick application of the Rational Root Theorem.
- ✓If you find one root, say x=a, you know (x-a) is a factor. Use polynomial long division to find the remaining factors, which is often more systematic than inspection.