Quadratic Functions and Equations in One Variable
How to Sketch a quadratic graph
Use this to sketch a parabola showing its shape, intercepts and turning point.
Before you start
- Deciding a parabola's direction from the sign of a
- Finding intercepts by substituting or solving
- Completing the square to find the turning point
When to use it
Use this to sketch a parabola showing its shape, intercepts and turning point.
The steps
- Look at the sign of a: positive opens upward, negative opens downward.
- Find the y-intercept by setting x = 0.
- Find the x-intercepts (roots) by solving the equation.
- Find the turning point by completing the square.
- Plot these features and draw a smooth curve through them.
Worked example
Sketch the graph of y = x² − 4x + 3, showing its intercepts and turning point.
- The coefficient of x² is a = 1, which is positive, so the parabola opens upward.
- Find the y-intercept by setting x = 0: y = 3, giving the point (0, 3).
- Find the x-intercepts by solving x² − 4x + 3 = 0, which factorises to (x − 1)(x − 3) = 0, so x = 1 and x = 3.
- Find the turning point by completing the square: x² − 4x + 3 = (x − 2)² − 1, so the turning point is (2, −1), a minimum.
- Plot (0, 3), (1, 0), (3, 0) and (2, −1), then draw a smooth U-shaped curve through them.
A second example, with a twist
Here a is negative, so the parabola opens downward and its turning point is a maximum instead of a minimum. Sketch the graph of y = −x² + 2x + 3, showing its intercepts and turning point.
- The coefficient of x² is a = −1, which is negative, so the parabola opens downward.
- Find the y-intercept by setting x = 0: y = 3, giving the point (0, 3).
- Find the x-intercepts by solving −x² + 2x + 3 = 0; multiplying by −1 gives x² − 2x − 3 = 0, which factorises to (x − 3)(x + 1) = 0, so x = 3 and x = −1.
- Find the turning point by completing the square: y = −(x² − 2x) + 3 = −((x − 1)² − 1) + 3 = −(x − 1)² + 4, so the turning point is (1, 4), a maximum.
- Plot (0, 3), (3, 0), (−1, 0) and (1, 4), then draw a smooth ∩-shaped curve through them.
Practise this in a KBAT problem
Frequently asked questions
Do I always need every feature to sketch?
For a good sketch you want the direction, the y-intercept, the x-intercepts if they exist, and the turning point. Together they fix the shape and position.
A sketch is not to scale, but it should show these key features clearly and in the right places relative to the axes.
What if the quadratic has no x-intercepts?
If the equation has no real roots, the curve never crosses the x-axis. You can tell because the discriminant b² − 4ac is negative.
Then rely on the direction, the y-intercept and the turning point to place the curve, it sits entirely above or entirely below the x-axis.
Is the turning point always halfway between the roots?
Yes, when the two roots exist, the turning point sits exactly midway between them because a parabola is symmetric. So its x-coordinate is the average of the roots.
Completing the square gives the same x-value, and also the y-value, which the midpoint alone does not tell you.