Graphs of Quadratic Functions
This topic covers the key features of quadratic graphs (parabolas), focusing on how to identify intercepts and turning points from a graph and how to calculate them precisely using algebraic methods like completing the square.
Part of the ESAT Mathematics 1 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The graph of y = ax2 + bx + c is a parabola. It's U-shaped (has a minimum point) if a > 0, and n-shaped (has a maximum point) if a < 0.
- The 'roots' are the x-intercepts, where the graph crosses the x-axis. A quadratic can have 0, 1, or 2 real roots.y=0
- The y-intercept is where the graph crosses the y-axis. Its coordinates are always (0, c).x=0
- The turning point (vertex) is the minimum or maximum point of the parabola. Its x-coordinate lies on the line of symmetry.
- Completing the square changes the form to y = a(x - p)2 + q, which directly reveals the turning point's coordinates as (p, q).
Diagram
Formulae
y = a(x - p)2 + q This is the 'completed square' form of a quadratic. Use it to find the coordinates of the turning point, which are (p, q). The value 'q' represents the minimum or maximum value of the function.
Definitions
- Roots
- The x-values for which a function equals zero. On a graph, they are the points where the curve intersects the x-axis.
- Turning Point
- The point on a parabola where the gradient changes sign. It is the minimum point of a U-shaped curve or the maximum point of an n-shaped curve.
- Intercepts
- The points where a graph crosses the coordinate axes. The y-intercept occurs at x=0, and the x-intercepts (roots) occur at y=0.
Worked example
A quadratic function is given by f(x) = -2x2 - 8x - 5. By completing the square, find the coordinates of its turning point and state whether it is a maximum or a minimum.
- 1
First, factor out the coefficient of x2 from the first two terms:
f(x) = -2(x2 + 4x) - 5 - 2
Next, complete the square for the expression inside the brackets, x2 + 4x.
This becomes (x + 2)2 - 4.
- 3
Substitute this back into the function:
f(x) = -2[ (x + 2)2 - 4 ] - 5 - 4
Expand the outer brackets, remembering to multiply both terms inside by -2:
f(x) = -2(x + 2)2 + 8 - 5 - 5
Simplify to get the final form:
f(x) = -2(x + 2)2 + 3 - 6
From the form a(x - p)2 + q, we can see p = -2 and q = 3.
The turning point is at (-2, 3).
- 7
Since the coefficient of x2 is -2 (which is less than 0), the parabola is n-shaped, so the turning point is a maximum.
Answer: The turning point is at (-2, 3) and it is a maximum.
Common mistakes
- ×Sign errors when finding the turning point from the completed square form. For y = (x - p)2 + q, the x-coordinate of the turning point is p, not -p.
- ×Forgetting to multiply the constant term after completing the square. In a[ (x+k)2 - k2 ], the second term becomes -ak2 when the bracket is expanded, not -k2.
- ×Arithmetic errors with fractions or negative numbers are common. Take care when halving odd coefficients or squaring negative values.
No-calculator tips
- ✓If you can spot the roots by factorising, the x-coordinate of the turning point is simply their average. For roots at x=1 and x=5, the turning point is at x = (1+5)/2 = 3.
- ✓The value of 'c' in y = ax2 + bx + c is always the y-intercept. This is a quick point to find to help sketch or interpret the graph.
- ✓Before starting a long calculation, check the sign of 'a'. This immediately tells you if you're looking for a minimum (a>0) or a maximum (a<0) and helps you self-check your answer.