Sketch the graph of a quadratic function, find the intercepts and the turning point, and use the line of symmetry

What this dot point is asking

SEAB wants you to sketch the graph of a quadratic function, find where it crosses the axes, locate its turning point and line of symmetry, and recognise how the sign of the leading coefficient sets the shape. A confident sketch turns many algebraic questions into something you can read off a picture.

The answer

The parabola

The graph of $y = ax^2 + bx + c$ is a parabola, a smooth symmetric U-shaped curve. If $a > 0$ the parabola opens upward and has a minimum point; if $a < 0$ it opens downward and has a maximum point. The larger $|a|$, the narrower the curve.

The intercepts

The curve crosses the $y$-axis where $x = 0$, giving the point $(0, c)$. It crosses the $x$-axis where $y = 0$, found by solving the quadratic $ax^2 + bx + c = 0$. There may be two, one, or no $x$-intercepts depending on the discriminant.

The turning point

The turning point (vertex) is the lowest point of an upward parabola or the highest point of a downward one. Completing the square to write the function as $a(x - h)^2 + k$ shows the turning point directly at $(h, k)$. The minimum or maximum value of $y$ is $k$.

The line of symmetry

A parabola is symmetric about a vertical line through its turning point, with equation $x = h$. This line of symmetry sits exactly halfway between the two $x$-intercepts when they exist, which is a quick way to find $h$.

Finding the turning point without completing the square

When the $x$-intercepts are known, there is a quicker route to the turning point than completing the square: because a parabola is symmetric, the line of symmetry sits exactly midway between the two roots. Average the roots to get the $x$-coordinate of the vertex, then substitute that value into the function to find the minimum or maximum $y$. For $y = x^2 - 2x - 3$ with roots $-1$ and $3$, the axis is $x = \tfrac{-1 + 3}{2} = 1$, and substituting gives $y = -4$, so the vertex is $(1, -4)$. Using symmetry of the roots is the fastest method whenever the quadratic factorises.

Reading the discriminant from the graph

The number of times the parabola crosses the $x$-axis matches the discriminant $b^2 - 4ac$ of the quadratic. Two crossings mean a positive discriminant, the curve just touching the axis at its vertex means a zero discriminant (a repeated root), and the curve missing the axis entirely means a negative discriminant with no real roots. So a parabola whose vertex sits above the $x$-axis while opening upward has no real roots. Linking the picture to the discriminant lets you predict, before solving, how many $x$-intercepts to expect and serves as a check on your algebra.

Examples in context

Example 1. Projectile paths. The height of a thrown ball against time traces a downward parabola, with the turning point giving the maximum height and the $x$-intercepts giving launch and landing times. Reading the vertex answers how high and the intercepts answer when.

Example 2. Maximising area. For a fixed perimeter, the area of a rectangle as a function of one side is a downward parabola, and its turning point gives the dimensions of greatest area. The graph makes the optimum visible at a glance.

Try this

Q1. State whether $y = -2x^2 + 3x + 1$ has a maximum or minimum, and why. [1 mark]

• Cue. The coefficient of $x^2$ is negative, so the parabola opens downward and has a maximum.

Q2. Find the $y$-intercept of $y = x^2 + 5x - 6$. [1 mark]

• Cue. Set $x = 0$: $y = -6$, the point $(0, -6)$.

Q3. Find the $x$-intercepts of $y = x^2 - x - 12$. [2 marks]

• Cue. Factorise: $(x - 4)(x + 3) = 0$, so the curve crosses at $(4, 0)$ and $(-3, 0)$.