Quadratic Graphs

Coordinate Geometry and Graphs

From the completed square form

Rewriting the equation as y = a(x − p)² + q puts the turning point at (p, q), and the line of symmetry at x = p
Watch the sign: it is the value that makes the bracket zero, so y = (x + 3)² − 1 turns at (−3, −1)
The number a changes how steep the curve is but not where it turns
A positive a makes the turning point a minimum and a negative a makes it a maximum
Finding the turning point this way is explicitly required by this specification

Exam tip

The sign flips coming out of the bracket: y = (x + 3)² − 1 turns at x = −3, not x = 3. Write the completed square form on its own line before reading the coordinates off it, because that line carries a mark of its own.

Worked example

An n-shaped curve and its maximum

Find the turning point of y = 12 + 4x − x² and say whether it is a maximum or a minimum.

Solution:

Rewrite in the usual order: y = −x² + 4x + 12
Take out the −1 from the first two terms: y = −[x² − 4x] + 12
Half of −4 is −2, so x² − 4x is (x − 2)² − 4
Substitute back: y = −[(x − 2)² − 4] + 12
Multiply through the bracket: y = −(x − 2)² + 4 + 12, so y = −(x − 2)² + 16
The turning point is (2, 16), and since the number in front of the squared bracket is negative it is a maximum
Check by substituting x = 2 into the original: 12 + 8 − 4 = 16

From the two axis crossings

Where the curve does cross the x-axis twice, the turning point lies half way between the crossings
Average the two crossing values to get the x-coordinate, then substitute it back to get the y-coordinate
For y = x² − 2x − 8 the crossings are at 4 and −2, so x = 1 and y = 1 − 2 − 8 = −9, giving (1, −9)