Quadratic Vertex & Axis

\( x_{\text{vertex}}= -\tfrac{b}{2a},\quad y_{\text{vertex}}= f\!\left(-\tfrac{b}{2a}\right),\quad \text{axis}:\;x=-\tfrac{b}{2a} \)
Algebra GCSE
Question 4 of 20
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\( Find the axis of symmetry of y = x^2 - 6x + 5 \)

Tips: use ^ for powers, sqrt() for roots, and type pi for π.
Hint (H)
\( Use x = -b/(2a) \)

Explanation

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Statement

A quadratic function is any function that can be written in the form \( f(x) = ax^2 + bx + c \), where \(a \neq 0\). Its graph is a parabola, which may open upwards (\(a > 0\)) or downwards (\(a < 0\)). The most important point on a parabola is its vertex, which represents the maximum point if the parabola opens downwards, or the minimum point if the parabola opens upwards. The vertical line passing through this point is called the axis of symmetry.

\[ x_{\text{vertex}} = -\frac{b}{2a}, \quad y_{\text{vertex}} = f\!\left(-\frac{b}{2a}\right), \quad \text{axis: } x = -\frac{b}{2a}. \]

Why it’s true

  • The axis of symmetry divides the parabola into two mirror halves. At this line, the slope of the tangent is zero, giving the extremum point.
  • By completing the square or differentiating, we can show that the x-coordinate of the vertex is always \(-\tfrac{b}{2a}\).
  • Once the x-coordinate is known, substituting it into the function gives the y-coordinate of the vertex.

Recipe (how to use it)

  1. Identify \(a\), \(b\), and \(c\) in the quadratic \(ax^2 + bx + c\).
  2. Calculate \(x_{\text{vertex}} = -\tfrac{b}{2a}\).
  3. Substitute this value into the quadratic to find \(y_{\text{vertex}}\).
  4. Write the axis of symmetry as \(x = x_{\text{vertex}}\).

Spotting it

Whenever you are asked for the maximum/minimum value of a quadratic or the axis of symmetry of a parabola, you should use this formula.

Common pairings

  • Finding the range of a quadratic function.
  • Solving optimisation problems in contexts such as area and revenue.
  • Sketching quadratic graphs accurately.

Mini examples

  1. Given: \(y = 2x^2 - 4x + 1\). Find: the vertex. Answer: \((1, -1)\).
  2. Given: \(y = -x^2 + 6x - 5\). Find: axis of symmetry. Answer: \(x = 3\).

Pitfalls

  • Forgetting to divide by \(2a\) instead of just 2.
  • Using the wrong sign for \(b\); remember it is \(-b\).
  • Not substituting back to find \(y_{\text{vertex}}\).
  • Mixing up maximum and minimum; check the sign of \(a\).

Exam strategy

  • Always write down \(a\), \(b\), and \(c\) before applying the formula.
  • Show substitution clearly to avoid sign errors.
  • If time allows, sketch the parabola quickly to check your answer makes sense.

Summary

The vertex and axis of symmetry give the turning point and line of reflection for a quadratic. The formula \(x = -\tfrac{b}{2a}\) is a quick way to find them, and it underpins both graph sketching and real-world optimisation problems. Mastery of this tool makes quadratic questions much more manageable in GCSE exams.

Worked examples

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  1. \( Find the vertex of y = x^2 + 4x + 3 \)
    1. \( a = 1, b = 4 \)
    2. \( x_vertex = -b/(2a) = -4/(2*1) = -2 \)
    3. \( y_vertex = f(-2) = (-2)^2 + 4(-2) + 3 = -1 \)
    Answer: (-2, -1)
  2. \( Find the axis of symmetry for y = 2x^2 - 8x + 5 \)
    1. \( a = 2, b = -8 \)
    2. \( x = -(-8)/(2*2) = 8/4 = 2 \)
    Answer: \( x = 2 \)
  3. \( Find the turning point of y = -x^2 + 6x - 5 \)
    1. \( a = -1, b = 6 \)
    2. \( x = -6/(2*-1) = 3 \)
    3. \( y = -(3)^2 + 6*3 - 5 = 4 \)
    Answer: (3, 4)
  4. \( Determine the minimum value of y = 2x^2 - 4x + 7 \)
    1. \( a = 2, b = -4 \)
    2. \( x = -(-4)/(2*2) = 1 \)
    3. \( y = 2(1)^2 - 4(1) + 7 = 5 \)
    Answer: \( Minimum value 5 at x = 1 \)
  5. \( Find the axis of symmetry for y = 3x^2 + 12x + 1 \)
    1. \( a = 3, b = 12 \)
    2. \( x = -12/(2*3) = -2 \)
    Answer: \( x = -2 \)
  6. \( Find the vertex of y = -2x^2 + 4x + 1 \)
    1. \( a = -2, b = 4 \)
    2. \( x = -4/(2*-2) = 1 \)
    3. \( y = -2(1)^2 + 4(1) + 1 = 3 \)
    Answer: (1, 3)
  7. \( Find the turning point of y = 0.5x^2 - x - 6 \)
    1. \( a = 0.5, b = -1 \)
    2. \( x = -(-1)/(2*0.5) = 1 \)
    3. \( y = 0.5(1)^2 - 1 - 6 = -6.5 \)
    Answer: (1, -6.5)
  8. \( Find the axis of symmetry for y = -3x^2 - 6x + 2 \)
    1. \( a = -3, b = -6 \)
    2. \( x = -(-6)/(2*-3) = -1 \)
    Answer: \( x = -1 \)
  9. \( Find the maximum value of y = -x^2 + 2x + 3 \)
    1. \( a = -1, b = 2 \)
    2. \( x = -2/(2*-1) = 1 \)
    3. \( y = -(1)^2 + 2*1 + 3 = 4 \)
    Answer: \( Maximum value 4 at x = 1 \)
  10. \( Find the vertex of y = 4x^2 - 12x + 11 \)
    1. \( a = 4, b = -12 \)
    2. \( x = -(-12)/(2*4) = 1.5 \)
    3. \( y = 4(1.5)^2 - 12(1.5) + 11 = 2 \)
    Answer: (1.5, 2)