TheIBAxolotl
Three forms, three views
A quadratic is any function \(f(x) = ax^2 + bx + c\) with \(a \ne 0\). Its graph is a parabola: it opens upward if \(a > 0\) and downward if \(a < 0\). The same quadratic can be written three ways, and each one reveals something without any working:
| Form | Shows directly |
|---|---|
| Standard \(ax^2 + bx + c\) | \(y\)-intercept \(c\) |
| Factorised \(a(x - p)(x - q)\) | roots \(p\) and \(q\) |
| Vertex \(a(x - h)^2 + k\) | vertex \((h, k)\) |
Axis of symmetry and vertex
A parabola is symmetric about the vertical line through its vertex. From standard form the axis is
\[x = -\frac{b}{2a}.\]
Substitute that \(x\) back into \(f\) to get the vertex's \(y\)-coordinate. From factorised form the axis is halfway between the roots, \(x = \frac{p + q}{2}\).
Completing the square
This converts standard form into vertex form. The trick is that \((x + m)^2 = x^2 + 2mx + m^2\), so a term in \(x\) tells you the \(m\) you need.
Worked example
Write \(f(x) = 2x^2 - 12x + 10\) in vertex form and sketch it.
- Factor out the \(a\) from the \(x\) terms: \(2(x^2 - 6x) + 10\).
- Complete the square inside: \(x^2 - 6x = (x - 3)^2 - 9\).
- So \(f(x) = 2\big((x - 3)^2 - 9\big) + 10 = 2(x - 3)^2 - 8\).
Now read the features:
- Vertex \((3, -8)\), axis \(x = 3\), opens upward (\(a = 2 > 0\)), so \(-8\) is the minimum value.
- \(y\)-intercept: \(f(0) = 10\).
- Roots: \(2(x^2 - 6x + 5) = 2(x - 1)(x - 5) = 0\), so \(x = 1\) and \(x = 5\), which are symmetric about \(x = 3\) as expected.
Check: \(f(3) = 18 - 36 + 10 = -8\). A sketch shows the U-shape through \((0, 10)\), \((1, 0)\), \((5, 0)\) with its lowest point at \((3, -8)\).
When it does not factorise
Use the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]
The expression under the root, \(b^2 - 4ac\), decides how many roots there are: positive means two, zero means one (the vertex touches the axis), negative means none.
Common mistakes
- Forgetting to multiply the \(-9\) by the outside factor 2 when completing the square.
- Reading the vertex of \(a(x - h)^2 + k\) as \((-h, k)\); the sign inside flips: \((x - 3)^2\) has \(h = 3\).
- Using \(x = \frac{b}{2a}\) without the minus sign.
- Assuming the vertex is a maximum without checking the sign of \(a\).
Exam tips
- Choose the form that answers the question; convert only if needed.
- Use symmetry: once you know one root and the axis, the other root is a reflection.
- In "find the quadratic" questions, start from the form that matches the given information (roots given, use factorised; vertex given, use vertex form) and use one extra point to find \(a\).