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# How to Quickly Find the Vertex of a Quadratic

Factorising and solving quadratics is probably the hardest topic for many GCSE students to grasp. It demands competent algebra skills and an understanding of graphing this particular type of function. Still, it is so important to master the tricks for solving them if you plan on studying Maths or Engineering at A-level and beyond. They come up everywhere!

This blog will cover the two main methods of solving a quadratic and then cover two bonus methods that allow you to find the inflexion point (vertex) of quadratics with larger coefficients without needing to simplify properly and factorise it first which can be time-consuming under exam pressure. These points show where the apex of the graph curve lies - so it occurs where the gradient is equal to zero and the function is at its minimum or maximum value.

The two main methods are:

## 1. Completing The Square

Example: Find the vertex / minimum turning point of the quadratic.

Solution: By completing the square, this becomes:

Giving the vertex:

## 2. Find The Roots Using The Symmetry of The Curve.

Example:

Factorise as standard to find the solution:

Therefore the roots where the curve crosses the x-axis are found when y = 0:

By symmetry, the vertex occurs at the midpoint of the roots, where:

and

vertex =

## 3. Quick Method 1

What happens if there are no roots and completing the square is more troublesome – for example:

A slick method is to use a part of the familiar quadratic formula to find the x-coordinate of the vertex rather than using the whole formula to find the roots:

Isolating this section and subbing in the coefficients for a and b gives:

The y-coordinate of the vertex is then obtained by evaluating the quadratic formula for y when:

Using a calculator, this gives you:

Hence the vertex is at:

## 4. Quick Method 2

The second method is similar but uses the fact that in any quadratic formula in the form:

...the extra number c simply shifts the simpler, related quadratic either up or down through a translation, but both parabolas will have the same x-coordinate for their vertices.

Hence, since the terms containing the variable:

....will always factorise, this method is similar to method 1 above. Using this method, we will find the vertex of:

First, solve to find the roots of:

By symmetry, the x-coordinate of the vertex is half-way between these, which is:

The y-coordinate is found by using the formula in method 3 as before:

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