How to sketch a quadratic graph — the steps
How do you sketch a quadratic graph?
Sketching a quadratic is a fixed routine that turns an algebraic expression into an accurate picture of its graph, without guessing point by point. The steps are always in the same order: first decide the shape from the sign of the x² coefficient, then find where the curve crosses the axes, and finally find the turning point — the minimum or maximum. In this guide we sketch several quadratics from start to finish.
Worked examples, step by step
Sketch the graph of y = x² − 4x + 3
- Shape: the coefficient of x² is 1, which is positive, so the parabola is U-shaped with a minimum
- y-intercept: substitute x = 0 to get y = 3 — the point (0,3)
- Roots: solve x² − 4x + 3 = 0, factorise to (x − 1)(x − 3) = 0, so x = 1 or x = 3 — the points (1,0) and (3,0)
- Turning point: by symmetry it is midway between the roots, at x = 2, and y = 4 − 8 + 3 = −1, giving (2,−1); the graph falls until x = 2 and rises afterwards
Find the turning point of y = x² − 6x + 5 without finding the roots
- Complete the square: x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4
- A square is never negative, so (x − 3)² − 4 is smallest when (x − 3)² = 0, that is when x = 3
- At x = 3 the value is −4, so the turning point is (3,−4), a minimum
- Check: substituting x = 3 into the original gives 9 − 18 + 5 = −4
Sketch y = −x² + 2x + 3
- Shape: the coefficient of x² is −1, which is negative, so the parabola is an upside-down U with a maximum
- y-intercept: x = 0 gives y = 3 — the point (0,3)
- Roots: −x² + 2x + 3 = 0 is the same as x² − 2x − 3 = 0, which factorises to (x − 3)(x + 1) = 0, so x = 3 or x = −1
- Turning point: midway between the roots, x = 1, and y = −1 + 2 + 3 = 4 — the maximum is (1,4)
Sketch y = x² + 2x + 5 and explain why it does not cross the x-axis
- Complete the square: x² + 2x + 5 = (x + 1)² + 4
- (x + 1)² is at least 0, so y is at least 4 — the graph never reaches y = 0, so there are no roots
- The turning point is (−1,4), a minimum, and the y-intercept is (0,5)
- Sketch: a U-shape sitting entirely above the x-axis, lowest at (−1,4)
Now try it yourself
Frequently asked questions
How do I know whether the parabola has a maximum or a minimum?
From the sign of the x² coefficient. Positive means the parabola opens upwards (U-shape) and the turning point is a minimum; negative means it opens downwards (an upside-down U) and the turning point is a maximum.
How do I find the turning point without completing the square?
Use symmetry: a parabola is symmetric about the vertical line through its turning point, so the x-coordinate of the turning point is midway between the two roots. Substitute that x back into the equation to get the y-coordinate. This works whenever the quadratic has real roots; the formula x = −b ÷ 2a works in every case.
What if the quadratic does not factorise?
Use the quadratic formula x = (−b ± √(b² − 4ac)) ÷ 2a to find the roots, or complete the square to find the turning point directly. If b² − 4ac is negative there are no real roots and the graph does not cross the x-axis at all.
What should a GCSE sketch actually show?
The correct shape, the y-intercept, the roots (if any) labelled on the x-axis, and the turning point labelled with its coordinates. A sketch is not a plot — it does not need to be to scale — but every key point must be in the right place relative to the others.
✍️ Written by the MathsUK teamChecked against the National Curriculum and GCSE specificationsLast updated: