Practice Maths

Quadratic Functions — Graphs and Features

Key Terms

A quadratic function has the form f(x) = ax² + bx + c, where a ≠ 0. Its graph is a parabola.
If a > 0 the parabola opens upward (concave up, minimum turning point). If a < 0 it opens downward (concave down, maximum turning point).
The axis of symmetry is the vertical line x = −b/(2a).
The turning point (vertex) lies on the axis of symmetry: x = −b/(2a), then substitute to find y.
The y-intercept is found by substituting x = 0: y = c.
The x-intercepts are found by solving ax² + bx + c = 0 (factorising, completing the square, or quadratic formula).
The discriminant Δ = b² − 4ac tells you how many x-intercepts: Δ > 0 (two), Δ = 0 (one, touching), Δ < 0 (none).
Vertex form
f(x) = a(x − h)² + k puts the vertex at (h, k) directly.
FeatureHow to find it
y-interceptSet x = 0 → y = c
Axis of symmetryx = −b/(2a)
Turning pointx = −b/(2a), then y = f(x)
x-interceptsSolve ax² + bx + c = 0 (discriminant first)
Number of x-interceptsΔ > 0: two; Δ = 0: one; Δ < 0: none

Example graph: y = x² − 2x − 3

Features: turning point (1, −4), y-intercept (0, −3), x-intercepts (−1, 0) and (3, 0), axis of symmetry x = 1.

x y −2 −1 1 2 3 4 2 1 −1 (−1, 0) (3, 0) y-int (0,−3) x = 1 TP (1,−4) y = x² − 2x − 3
Hot Tip Always find the axis of symmetry first — the turning point sits on it. Use x = −b/(2a), substitute back to get the y-coordinate. A parabola is symmetric about this axis, so x-intercepts (if they exist) are equal distances either side.

Worked Example 1 — Finding All Key Features

Question: For f(x) = x² − 2x − 3, find the axis of symmetry, turning point, y-intercept, and x-intercepts.

Step 1 (axis of symmetry): a = 1, b = −2. x = −(−2)/(2×1) = 1.

Step 2 (turning point): f(1) = 1 − 2 − 3 = −4. Turning point: (1, −4) — minimum (a > 0).

Step 3 (y-intercept): f(0) = −3. y-intercept: (0, −3).

Step 4 (x-intercepts): Solve x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0 → x = 3 or x = −1.

Answer: Axis: x = 1; TP: (1, −4) min; y-int: (0, −3); x-ints: (3, 0) and (−1, 0).

Worked Example 2 — Vertex Form and Transformations

Question: Write g(x) = 2x² − 8x + 5 in vertex form and state the turning point.

Step 1 (complete the square): g(x) = 2(x² − 4x) + 5

Step 2: Complete the square inside: x² − 4x = (x − 2)² − 4.

Step 3: g(x) = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3.

Answer: Vertex form: g(x) = 2(x − 2)² − 3. Turning point: (2, −3), minimum (a = 2 > 0).

Key Features and What They Really Mean

The vertex (turning point) is the most important feature of a parabola. If a > 0, the parabola opens upward and the vertex is the minimum value — the function cannot go below this point. If a < 0, it opens downward and the vertex is the maximum. This is why vertex form is so valuable: in y = a(x − h)² + k, the minimum (or maximum) value of the function is immediately visible as k.

The axis of symmetry is not merely a line — it is the mirror of the parabola. Every point on one side of the axis has a mirror image at the same height on the other side. This symmetry is why x-intercepts (when they exist) are always symmetric about the axis, and why the vertex x-coordinate is exactly halfway between the two x-intercepts.

Three Forms and What Each Reveals

Quadratics can be written in three equivalent forms, each making different features immediately visible:

Standard form y = ax² + bx + c: the y-intercept is simply c (set x = 0). The axis of symmetry is x = −b/(2a). This is the form you get after expanding, but it hides the vertex and x-intercepts.

Vertex (turning point) form y = a(x − h)² + k: the vertex (h, k) is instantly visible. To convert from standard form, complete the square. This form also makes transformations transparent: shift h right, k up, dilate/reflect by factor a.

Factored form y = a(x − p)(x − q): the x-intercepts p and q are immediately visible. Note the vertex x-coordinate is always halfway between p and q: x = (p + q)/2, which is consistent with x = −b/(2a).

A key skill is being able to convert between all three forms fluently. In exams, you may be given one form and need another. Know which information each form provides at a glance.

The Role of ‘a’: Dilation and Reflection

The coefficient a in y = ax² controls both the width and the orientation of the parabola. Understanding this precisely:

If |a| > 1, the parabola is narrower than y = x² — a dilation closer to the y-axis. If 0 < |a| < 1, it is wider. The vertical scale factor is |a|: at any given x-displacement from the axis, the height changes by a factor of |a| compared to y = x².

If a < 0, the parabola is reflected in the x-axis: every y-value is negated, turning a minimum into a maximum and flipping the curve upside down. This is a reflection, not just a rotation.

Finding the Vertex from Standard Form

Given y = ax² + bx + c, the axis of symmetry is x = −b/(2a). This formula comes from completing the square on the general form. Substitute this x-value back into the original equation to find the y-coordinate of the vertex.

Why is the formula x = −b/(2a) and not x = b/(2a)? Because when we complete the square: y = a(x + b/(2a))² + (c − b²/(4a)), the vertex is at x = −b/(2a). The negative sign is critical — one of the most common errors in parabola problems.

Alternative: if you have the x-intercepts p and q, the axis is at x = (p + q)/2 by symmetry. This can be faster than the formula.

Sketching Strategy: A Systematic Approach

A reliable sketch requires four steps in order: (1) Determine the sign of a for direction. (2) Find the vertex using x = −b/(2a) and substitution, or completing the square. (3) Find the y-intercept (set x = 0). (4) Find x-intercepts: check Δ first — if Δ < 0 there are none; if Δ ≥ 0, solve. When there are no x-intercepts, the parabola is entirely above (a > 0) or below (a < 0) the x-axis. Label this clearly on your sketch.

A good sketch shows the shape clearly, labels all key points with coordinates, indicates the direction of opening, marks the axis of symmetry as a dashed line, and is large enough to read. Marks are awarded for correct features, not artistic quality.

Exam Tip: The axis of symmetry is x = −b/(2a), not x = b/(2a). This sign error is extremely common. A useful check: for y = x² − 6x + 5, the axis should be x = 3 (between the roots x = 1 and x = 5). Using −b/(2a) = −(−6)/(2×1) = 6/2 = 3 confirms this. If you accidentally use +b/(2a), you get x = −3, which is clearly wrong for this parabola.
Exam Tip: When Δ < 0 and there are no x-intercepts, do not leave your sketch with the parabola vaguely floating. Explicitly mark the vertex, draw the parabola clearly above (or below) the x-axis, and label it. If the question asks you to “sketch”, you must show the axis of symmetry, vertex coordinates, and y-intercept even when x-intercepts do not exist.

Mastery Practice

See Answers ➔

  1. For each quadratic, state: (i) whether the parabola opens up or down, (ii) the y-intercept, (iii) the axis of symmetry. Fluency

    1. (a) f(x) = x² + 4x − 5
    2. (b) g(x) = −2x² + 6x
    3. (c) h(x) = 3x² − 12x + 1
    4. (d) k(x) = −x² + 2x + 8

  2. Use the discriminant to determine the number of x-intercepts for each parabola. Fluency

    1. (a) y = x² − 5x + 4
    2. (b) y = x² − 4x + 4
    3. (c) y = 2x² + x + 3
    4. (d) y = −x² + 6x − 9

  3. Find the turning point (vertex) of each parabola and state whether it is a maximum or minimum. Fluency

    1. (a) y = x² − 6x + 2
    2. (b) y = −x² + 4x + 1
    3. (c) y = 2x² + 8x − 3
    4. (d) y = −3x² + 12x − 7

  4. Write each quadratic in vertex form f(x) = a(x − h)² + k by completing the square. State the turning point. Fluency

    1. (a) f(x) = x² + 6x + 7
    2. (b) f(x) = x² − 4x + 1
    3. (c) f(x) = 2x² − 8x + 9
    4. (d) f(x) = −x² + 2x + 5

  5. For each quadratic, find the axis of symmetry, turning point, y-intercept, and x-intercepts (if they exist). Fluency

    1. (a) y = x² + 2x − 8
    2. (b) y = −x² + 5x − 4
    3. (c) y = x² − 4x + 5 (no real x-intercepts)

  6. Finding the equation. Understanding

    Working backwards from features.
    1. (a) A parabola has x-intercepts at x = −2 and x = 4, and passes through (0, −16). Find its equation in the form y = ax² + bx + c.
    2. (b) A parabola has vertex (3, −2) and passes through (5, 6). Write it in vertex form and then expand to standard form.
    3. (c) A parabola opens downward, has axis of symmetry x = 1, and passes through (0, 3) and (2, 3). Find the maximum value.

  7. Transformations of parabolas. Understanding

    Starting from y = x². Describe fully the transformation(s) applied to y = x² to give each of the following parabolas, then state the turning point.
    1. (a) y = (x + 3)² − 1
    2. (b) y = −2(x − 1)² + 4
    3. (c) y = (x − 2)² (identify the y-intercept by expansion)

  8. Range and practical contexts. Understanding

    Interpreting parabolas.
    1. (a) State the range of f(x) = (x − 2)² + 3.
    2. (b) State the range of g(x) = −x² + 4x − 1.
    3. (c) A ball is thrown and its height (in metres) after t seconds is h(t) = −5t² + 20t + 1. Find the maximum height and the time it occurs.

  9. Intersection of a parabola and a line. Problem Solving

    Algebraic investigation. The parabola P has equation y = x² − 3x + 2 and the line L has equation y = x − 1.
    1. (a) Find the x-intercepts and turning point of P and sketch its graph (label all features).
    2. (b) Find the points of intersection of P and L by solving the system simultaneously.
    3. (c) For what values of k does the line y = x + k intersect P at exactly one point? What is the geometric meaning of this?

  10. Discriminant conditions. Problem Solving

    Parameter problem. Consider the quadratic f(x) = x² + kx + (k + 3) where k is a real constant.
    1. (a) Find the discriminant of f(x) in terms of k.
    2. (b) Find the values of k for which f(x) has no x-intercepts (i.e., the parabola lies entirely above the x-axis).
    3. (c) Find the value of k for which the parabola just touches the x-axis. State the turning point in each case.
    4. (d) If k = −4, write f(x) in vertex form and state all key features.