To learn Quadratic Graphs quickly and correctly, start with one clear picture: the graph of a quadratic function is a parabola, a curved shape whose turning point helps you understand its highest or lowest value. Rather than memorising a picture, learn how the equation reveals the curve’s direction, its intercepts and its symmetry. These eduKateSG Mathematics Quick Notes teach you how to build a value table, sketch a quadratic graph and check it against the algebra, with friendly worked examples.
- Recognise the form: y = ax² + bx + c with a ≠ 0.
- See the direction: a > 0 opens upwards; a < 0 opens downwards.
- Find the y-intercept: substitute x = 0.
- Find the x-intercepts: solve ax² + bx + c = 0 when real roots exist.
- Find the turning point: use symmetry or x = −b/(2a).
- Check the sketch: plot points on both sides of the symmetry line.
First, understand the basic shape. The simplest quadratic graph is y = x². Try x = −2, −1, 0, 1 and 2. Their y-values are 4, 1, 0, 1 and 4. The outputs match in pairs because (−2)² = 2² and (−1)² = 1². The lowest point is (0, 0), and the curve opens upward. If you change the equation to y = −x², each y-value changes sign, so the curve opens downward and its turning point is a maximum. A quadratic graph is curved because its rate of change is not constant.
Second, connect a graph with familiar algebra. For y = x² − 4x + 3, the y-intercept occurs when x = 0, so y = 3 and the graph passes through (0, 3). The x-intercepts occur when y = 0, so factorise x² − 4x + 3 = (x − 1)(x − 3) = 0. The roots are x = 1 and x = 3, giving intercepts (1, 0) and (3, 0). These are not three disconnected calculations. They describe three locations on the same curve.
Third, use symmetry to find the turning point. The two roots of the example are 1 and 3, and their midpoint is x = 2. Substitute x = 2 into y = x² − 4x + 3: y = 4 − 8 + 3 = −1. Thus the turning point is (2, −1), and the axis of symmetry is x = 2. Because the coefficient of x² is positive, the curve opens upward and its minimum y-value is −1. It is easier to sketch a reliable parabola when you have its key points and symmetry rather than connecting random dots.
Worked example: create a value table and draw the curve
Take y = x² − 4x + 3. Substitute x = 0, 1, 2, 3 and 4. The corresponding y-values are 3, 0, −1, 0, 3. Plot (0, 3), (1, 0), (2, −1), (3, 0) and (4, 3) with a consistent axis scale. Draw a smooth U-shaped curve passing through them, not a zigzag joining the points with straight segments. The pattern is symmetrical around x = 2, and both intercepts agree with the factorisation.
What the coefficient a tells you
The coefficient a in y = ax² + bx + c controls the opening direction. When a is positive, the turning point is a minimum. When a is negative, it is a maximum. Compared with y = x², a larger absolute value of a causes a narrower-looking curve on identically scaled axes, while a non-zero absolute value smaller than 1 causes a wider-looking curve. For instance, y = 2x² rises more rapidly than y = x² as |x| increases. Do not confuse the sign of a with the sign of the y-intercept, which comes from c.
How to find the turning point without factorising
For y = ax² + bx + c, the x-coordinate of the turning point is x = −b/(2a). Consider y = x² + 2x − 3. Here a = 1 and b = 2, so x = −2/2 = −1. Substitute: y = (−1)² + 2(−1) − 3 = 1 − 2 − 3 = −4. The turning point is (−1, −4). The graph crosses the x-axis at x = −3 and x = 1, which are equally far from −1. This is a useful independent check.
Completing the square: a graph in one expression
Rewrite x² − 4x + 3 as (x − 2)² − 1. Because a square is never negative for real x, the smallest possible value of (x − 2)² is zero, attained when x = 2. Therefore the minimum value is −1, and the turning point is (2, −1). The form y = a(x − h)² + k reveals the turning point (h, k). The sign of a determines whether the point is a minimum or maximum. You can use either completing the square or the −b/(2a) rule to check your result.
What if the graph never crosses the x-axis?
Consider y = x² + 1. Its minimum is at (0, 1), so the entire graph lies at or above y = 1. The equation x² + 1 = 0 has no real solutions; hence the graph has no real x-intercepts. A quadratic graph may cross the x-axis twice, just touch once, or never meet it. These cases correspond to two distinct real roots, one repeated real root and no real roots respectively. Avoid inventing x-intercepts when the curve remains above or below the axis.
Quick plotting checklist
- Write the equation and check whether a is positive or negative.
- Substitute x = 0 to find the y-intercept.
- Solve y = 0 for any real x-intercepts.
- Find the axis of symmetry and turning point.
- Create a short table with points on both sides of the axis.
- Choose sensible axis scales and draw one smooth curve.
- Check that paired points are symmetrical around the turning point’s vertical line.
Mistakes that make a good calculation produce a wrong graph
- Drawing straight line segments: a parabola is a smooth curve.
- Swapping x- and y-intercepts: use y = 0 for the x-axis and x = 0 for the y-axis.
- Assuming every quadratic has two x-intercepts: some have one or none.
- Forgetting the sign of a: it decides whether the curve opens upward or downward.
- Finding the axis but not the turning-point y-value: substitute the axis x-value back into the equation.
- Using uneven graph scales: a distorted grid can hide symmetry.
Mini-test: explain before calculating
- For y = x² − 6x + 8, find the x-intercepts.
- For the same graph, find the axis of symmetry and turning point.
- Find the y-intercept of y = −2x² + 3x + 5.
- Does y = x² + 4 cross the x-axis at any real value?
Answers and checking
1. x² − 6x + 8 = (x − 2)(x − 4), so the x-intercepts are (2, 0) and (4, 0). 2. The axis is x = 3; at x = 3, y = 9 − 18 + 8 = −1, so the minimum is (3, −1). 3. With x = 0, y = 5, so the y-intercept is (0, 5). 4. No: x² + 4 is always at least 4 for real x.
Your ten-minute quadratic graph routine
- Minutes 1–2: recognise the sign of a and choose the opening direction.
- Minutes 3–5: find roots, the y-intercept and the turning point.
- Minutes 6–8: complete a small table and sketch a symmetrical curve.
- Minute 9: check two table values on opposite sides of the symmetry line.
- Minute 10: explain what each intercept and the turning point means.
If you can rebuild a curve from the equation and explain why its points belong where they do, you have learned the graph rather than simply copied its shape.
Connect this lesson to the next Quick Notes
Continue with Quadratic Equations · Coordinate Graphs · Algebraic Expansion and Factorisation · Functions and Function Notation. Keep each study session small: explain, solve, check and revisit tomorrow.
For the wider learning approach, see Secondary 1 Mathematics Tutor Clementi | Small Groups Tutorials. The original reference article remains unchanged.