Algebra · GCSE Maths
Quadratic graphs
GCSE Maths quadratic graphs: parabola shape, roots, y-intercept, turning point from completing the square, and sketching y = ax² + bx + c.
a > 0 is a U; a < 0 is an n. Roots are x-intercepts. Turning point from completing the square or −b/(2a).
The important bits
What you need to know
- 1
The graph of y = ax² + bx + c is a parabola. If a > 0 it opens upward (U); if a < 0 it opens downward (n).
- 2
The y-intercept is c, because when x = 0, y = c. Plot (0, c) first on every sketch.
- 3
Roots (x-intercepts) solve ax² + bx + c = 0 by factorising or the quadratic formula. A repeated root touches the axis; no real roots mean the curve stays above or below the axis.
- 4
The line of symmetry is x = −b/(2a). The turning point lies on this vertical line.
- 5
Completing the square gives turning point (p, q) from y = a(x − p)² + q. Minimum if a > 0, maximum if a < 0.
- 6
To sketch: mark intercepts, symmetry line, turning point, and draw a smooth curve — not straight segments.
- 7
The discriminant b² − 4ac tells how many x-intercepts: positive two, zero one (tangent), negative none.
- 8
Quadratic and linear graphs intersect where ax² + bx + c = mx + d; solve the resulting quadratic for intersection x-values.
Quotations worth analysing
Short evidence. Real method.
“x = −b/(2a)”
The turning point x-coordinate. Students who use +b/(2a) mirror the parabola the wrong way. Pair with y by substituting back.
“y = a(x − p)² + q”
Turning point (p, q). The sign in (x − p) is opposite to the coordinate: (x + 3)² means p = −3.
“Smooth parabola, not a V with a corner”
Join points with a curve. A sharp corner at the vertex suggests misunderstanding — quadratics are smooth everywhere.
Go deeper
From equation to sketch
Sketch y = x² − 4x + 3. y-intercept: (0, 3). Factorise: (x − 1)(x − 3) = 0, roots x = 1 and x = 3. Symmetry line x = (1 + 3)/2 = 2. At x = 2, y = 4 − 8 + 3 = −1, turning point (2, −1). Plot (0, 3), (1, 0), (3, 0), (2, −1), draw smooth U. For y = −x² + 4, y-intercept 4, roots ±2, maximum at (0, 4) — inverted n. Negative a flips the U. If roots are irrational, use the formula or symmetry: x = −b/(2a) for vertex x, then substitute.
Go deeper
Completing the square on the graph
y = 2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3. Turning point (2, −3), minimum because a = 2 > 0. Line of symmetry x = 2. Compare to y = 2x² − 8x + 5: b = −8, −b/(2a) = 8/4 = 2. Roots from 2(x − 2)² = 3, (x − 2)² = 1.5, x = 2 ± √1.5. Discriminant (−8)² − 4×2×5 = 64 − 40 = 24 > 0, two roots. Sketch shows vertex below the axis for this example.
Go deeper
Intersection with a line
y = x² and y = 2x + 3 meet when x² = 2x + 3, so x² − 2x − 3 = 0, (x − 3)(x + 1) = 0, x = 3 or x = −1. y-values: 9 and 1. Points (3, 9) and (−1, 1). Graphically: a line cutting a parabola twice. If the discriminant is zero, tangent; negative, no intersection. Exam questions may ask how many solutions without sketching: solve and interpret D. y = x² + 1 and y = x have x² − x + 1 = 0, D = 1 − 4 = −3, no real intersections — parabola sits above the line.
See the idea in action
Sketch y = x² − 6x + 5, labelling intercepts and the turning point. y-intercept: (0, 5). Factorise: (x − 1)(x − 5) = 0, roots x = 1 and x = 5. Symmetry: x = 3. At x = 3, y = 9 − 18 + 5 = −4. Turning point (3, −4). Sketch smooth U through (0, 5), (1, 0), (3, −4), (5, 0).
Exam technique
Turn knowledge into marks
Label roots, y-intercept and turning point on sketches. Use −b/(2a) for the symmetry line even when you factorise — it checks your vertex.
Common mistakes
Do not give these marks away
- 01
Drawing a V-shape with a sharp corner instead of a smooth parabola.
- 02
Confusing y-intercept c with the turning point y-coordinate.
- 03
Using +b/(2a) for the axis of symmetry instead of −b/(2a).
For y = 2x² − 8x + 3, the x-coordinate of the turning point is
A2
B−2
C4
D8
Show the answer
2. x = −b/(2a) = −(−8)/(2×2) = 8/4 = 2. −2 comes from sign error. 4 might be from b/(2a) without the minus.
Quick questions
If this is the bit you searched
How do I find the turning point quickly?
x = −b/(2a), substitute for y. Or complete the square to y = a(x − p)² + q, giving vertex (p, q).
What if the quadratic does not factorise?
Use the formula for roots, or the discriminant to count roots. The turning point still comes from −b/(2a).
Why is the y-intercept always c?
Substitute x = 0 in y = ax² + bx + c. Only c remains. The x-intercepts come from solving ax² + bx + c = 0.
How does a affect the graph?
Sign of a decides U versus n. Larger |a| makes the parabola narrower (steeper). Smaller |a| makes it wider.