{
  "title": "Completing the Square — {{ class_or_section }}",
  "session": "{{ session_id, e.g. ALG-2.4 }}",
  "concepts": ["quadratic functions", "vertex form", "completing the square"],
  "slides": [
    {
      "heading": "Why not just memorise the formula?",
      "blocks": [
        {
          "md": "A quadratic $ax^2 + bx + c$ hides its own shape. The quadratic formula spits out the roots, but it tells you nothing about **where the curve turns** or **how far it reaches down**. Completing the square rewrites the same expression so the vertex is simply *there to read*."
        },
        {
          "md": "By the end you will turn any $x^2 + bx + c$ into $\\left(x + h\\right)^2 + k$ in one line, and read the vertex $(-h,\\ k)$ straight off it.",
          "fragment": true
        }
      ],
      "speaker_notes": "Open by asking the class what the quadratic formula does NOT tell them. Target answer: the vertex. Hold the fragment until someone says it. Replace {{ class_or_section }} in the title with your period."
    },
    {
      "heading": "The move: halve, square, balance",
      "blocks": [
        {
          "md": "The identity underneath everything today:"
        },
        {
          "md": "$$x^2 + bx = \\left(x + \\tfrac{b}{2}\\right)^2 - \\left(\\tfrac{b}{2}\\right)^2$$"
        },
        {
          "md": "Take **half** of the coefficient of $x$, **square** it, then **add and subtract** it so the value is unchanged. The first three terms fold into a perfect square; the leftover $-\\left(\\tfrac{b}{2}\\right)^2$ is the price you pay.",
          "fragment": true
        }
      ],
      "speaker_notes": "Say the three verbs out loud — halve, square, balance — and have the class chant them back. This is the whole lesson in three words."
    },
    {
      "heading": "Worked example",
      "blocks": [
        {
          "md": "Put $f(x) = x^2 + 6x + 1$ into vertex form and state its minimum."
        },
        {
          "md": "Half of $6$ is $3$; $3^2 = 9$. Add and subtract $9$:\n\n$$x^2 + 6x + 1 = \\left(x^2 + 6x + 9\\right) - 9 + 1 = \\left(x + 3\\right)^2 - 8$$\n\nThe vertex is $(-3,\\ -8)$, so the minimum value of $f$ is $-8$, reached at $x = -3$.",
          "solution": true
        }
      ],
      "speaker_notes": "Work the left column live; reveal the solution block only after two or three students have committed to a value for the constant. Common slip: forgetting to carry the original $+1$."
    },
    {
      "heading": "Your turn",
      "blocks": [
        {
          "md": "Rewrite $g(x) = x^2 - {{ even_coefficient, e.g. 10 }}\\,x + {{ constant, e.g. 7 }}$ in the form $\\left(x + h\\right)^2 + k$ and give its vertex."
        },
        {
          "md": "Check: your $h$ should be $-\\tfrac{1}{2}$ of the $x$-coefficient, and $k$ is what is left after you subtract $h^2$. Swap the two placeholder numbers for the ones your class is working with.",
          "fragment": true
        }
      ],
      "speaker_notes": "Choose an even middle coefficient the first time through so the halving stays whole — fractions are a separate lesson. Circulate while the fragment is hidden."
    }
  ]
}