Exploring Quadratics in GCSE Maths

posted in: Uncategorized | 0

GCSE maths requires students to be able to solve quadratic equations. Quadratic equations are of the form

ax^2+bx+c =0,

where a,b,c are all constants. Deeper insight into quadratic equations can be gained by thinking about what a general quadratic looks like when drawn on a graph. The quadratic curve  y=ax^2+bx+c has a shape known as a parabola. To take a concrete example, consider the following figure.

The equation of this curve is y=x^2-4x+3 and if we set y=0 then we find a standard quadratic equation problem 0=x^2-4x+3. But we can see from the graph that setting y=0 is equivalent to looking at the  x-axis, so a quadratic equation is equivalent to the problem of finding the two x values where the parabola crosses the x-axis.

For GCSE Maths, students are taught three different ways to solve the quadratic equation: the quadratic formula, factorization and completing the square. In our example, the quadratic factorizes easily into

 x^2-4x+3 = (x-1)(x-3),

which gives solutions x=1 and x=3. Looking at the graph confirms this. The parabola crosses the x-axis at these points.

If the parabola had not crossed the axis at integer values, then there would not be a straightforward factorization, and the quadratic formula

 x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

would be the best way to find the roots of this equation.

Completing the square leads to another interesting way to consider this problem. For our example, completing the square gives


There is a nice way to interpret this by thinking in terms of functions. Consider the following figure.

Let us suppose that y=f(x)=x^2, i.e. the function f takes its argument and outputs the square of whatever comes in. The graph of y=x^2 is coloured red in the figure and is symmetric about the y-axis as expected. We know from function theory that if we take the graph of a function f(x) and shift it to the right by a constant amount a, that it can be represented by f(x-a). Similarly if we shift it to the left on the graph by the same amount, it would be written f(x+a).

The green parabola in the figure can be obtained from the red parabola simply by shifting the red one to the right by a constant value of 2. This means its curve can be described by the equation y=f(x-2)=(x-2)^2.

Similarly the blue parabola (our original example), can be obtained from the green one by shifting vertically downwards by a constant value of 1. In other words, this curve is given by y=f(x-2)-1=(x-2)^2-1, the same result that we get by completing the square of our original quadratic.

So, completing the square shows us how far to shift the quadratic y=x^2 in each direction to map it on to our particular example. Another way of saying this is that completing the square gives us the minimum point (or maximum point if the parabola is upside-down) of the parabola.

If we want to sketch a parabola, it helps to know certain key points on the graph. The point where the parabola cuts the y-axis is important and is given by the constant at the end of the quadratic expression. The two points where the parabola cuts the x-axis are also key, and are given by factorizing the quadratic or applying the quadratic formula. The minimum (or maximum) point is found by completing the square.

Leave a Reply