The Quadratic Function
1. Three useful forms of a quadratic
Switch form deliberately instead of forcing every problem through the quadratic formula.
Expanded
\[y=ax^2+bx+c\]Factorised
\[y=a(x-r_1)(x-r_2)\]Completed square
\[y=a(x-h)^2+k\]Quadratic formula
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]Each form answers a different exam question quickly: expanded form exposes coefficients and the discriminant; factorised form exposes roots; completed-square form exposes the vertex and range.
Worked example — complete the square
\[2x^2-12x+7=2(x^2-6x)+7=2[(x-3)^2-9]+7\] \[=2(x-3)^2-11.\]Hence the minimum value is \(-11\) at \(x=3\).
2. Discriminant and nature of roots
The discriminant is the fastest bridge between algebra, geometry and parameter conditions.
Discriminant
\[\Delta=b^2-4ac\]Two distinct real roots
\[\Delta>0\]Repeated real root / tangency
\[\Delta=0\]No real roots
\[\Delta<0\]Parameter/tangency pattern
For \(x^2+kx+9=0\) to have equal roots: \[k^2-36=0\Rightarrow k=\pm6.\]Exam tip. When a line is tangent to a parabola, substitute the line into the quadratic and set the resulting discriminant equal to zero.
3. Functions of roots \(\alpha,\beta\)
Many long-looking root questions collapse to sum/product algebra.
Vieta
\[\alpha+\beta=-\frac ba,\qquad \alpha\beta=\frac ca\]Square sum
\[\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\]Reciprocal roots
\[\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}\]Equation from roots
\[x^2-(\text{sum})x+(\text{product})=0\]Worked example — new equation from transformed roots
If \(\alpha,\beta\) are roots of \(2x^2-5x+1=0\), then \[\alpha+\beta=\frac52,\qquad \alpha\beta=\frac12.\] For roots \(\alpha+1,\beta+1\): \[S=\frac92,\qquad P=\alpha\beta+(\alpha+\beta)+1=4.\] So \(x^2-\frac92x+4=0\), or \(2x^2-9x+8=0\).Common mistake. Do not solve for \(\alpha\) and \(\beta\) individually unless the question forces you to. Symmetric expressions are usually designed for Vieta.
4. Parameter and range problems
Interpret verbal conditions before calculating; the condition decides the tool.
- Rearrange into a quadratic in the target variable.
- Identify what the wording implies: real roots, equal roots, no roots, positive roots, or a vertex bound.
- Translate that condition into a discriminant, Vieta or completed-square condition.
- Solve the resulting inequality/equation in the parameter and check endpoints.