A quadratic is not just an equation to solve. It is a function whose algebraic form, roots, turning point and graph all describe the same structure.
This guide connects the three most useful forms of a quadratic and shows how to choose the representation that exposes the information a question needs.
1. Three forms, one quadratic
Consider y=x²−6x+5. In expanded form, the coefficients are visible. Factorising gives y=(x−1)(x−5), exposing roots x=1 and x=5. Completing the square gives y=(x−3)²−4, exposing turning point (3,−4).
These expressions are equal. A strong student changes form deliberately rather than treating each as a different chapter.
2. Axis of symmetry
For y=a(x−h)²+k, the axis of symmetry is x=h and the turning point is (h,k). In expanded form y=ax²+bx+c, the axis is x=−b/(2a).
For x²−6x+5, x=−(−6)/2=3, agreeing with completed-square form.
3. Maximum or minimum?
If a>0, the parabola opens upward and the turning point is a minimum. If a<0, it opens downward and the turning point is a maximum.
Thus y=−2(x+1)²+7 has maximum value 7 at x=−1.
4. Roots are graph intersections
Solving ax²+bx+c=0 finds the x-coordinates where the graph y=ax²+bx+c meets the horizontal axis. Two distinct real roots mean two intersections; one repeated real root means tangency; no real roots means no horizontal-axis intersection.
5. Worked example: reconstruct from roots
A quadratic has roots 2 and −3 and passes through (0,−12). Write its equation.
Start with y=A(x−2)(x+3). At x=0, −12=A(−2)(3)=−6A, so A=2. Therefore y=2(x−2)(x+3).
The roots determine factors, but not the vertical scale. The extra point determines A.
6. Worked example: reconstruct from a turning point
A quadratic has turning point (4,−5) and passes through (2,3). Write its equation.
Use y=A(x−4)²−5. Substitute (2,3): 3=4A−5, so A=2. Hence y=2(x−4)²−5.
7. Symmetry predicts paired points
If a parabola has axis x=3 and contains (1,7), it also contains (5,7), because these x-values lie equally far from the axis.
This provides a graph check and can reduce unnecessary substitution.
8. Intersections with another graph
To find intersections of y=x²−2x−3 and y=x+1, equate outputs: x²−2x−3=x+1. This gives x²−3x−4=0=(x−4)(x+1), so x=4 or −1.
Substitute into either equation to obtain intersection points (4,5) and (−1,0).
9. Common mistakes
- Calling the constant term c the vertical coordinate of the turning point.
- Finding roots and forgetting that they are x-coordinates.
- Using the sign inside (x−h)² as the turning-point x-coordinate without reversing it.
- Assuming roots determine a unique quadratic without a scale condition.
10. Practice
- Find the turning point of y=(x+3)²−8.
- Find the roots of y=x²−7x+12.
- State the axis of symmetry of y=2x²−12x+5.
- A quadratic has roots −1 and 4 and passes through (0,8). Find its equation.
- Find the intersections of y=x² and y=2x+3.
11. Answers
- (−3,−8).
- x=3 and x=4.
- x=3.
- y=−2(x+1)(x−4).
- x²=2x+3 gives (x−3)(x+1)=0; intersections (3,9) and (−1,1).
Continue through the Additional Mathematics Hub and the existing Quadratic Functions guide.