Graphs
Grade 5-7Quadratic Graphs
A quadratic graph always makes the same U-shaped curve, called a parabola, and the two features examiners ask about most are the roots, where the curve crosses the x-axis, and the turning point, its minimum or maximum. This lesson covers substituting values into a quadratic, reading roots and the turning point from a graph, finding roots by factorising, and finding the turning point by completing the square.
Written by Asad, Co-Founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Substituting values into a quadratic
To generate values for a quadratic graph, substitute each x value into the equation and work through the arithmetic carefully, remembering that squaring a negative number always gives a positive result.
For y = x² + 2, when x = -2: y = (-2)² + 2 = 4 + 2 = 6.
Reading roots and the turning point from a graph
The roots of a quadratic are the x-values where the curve crosses the x-axis, where y = 0. The turning point is the single lowest point (or highest, if the curve is upside down) on the curve, and sits exactly halfway between the two roots.
A graph of y = x² - 4x crosses the x-axis at x = 0 and x = 4, and its turning point is at (2, -4), halfway between the roots.
Finding roots by factorising
To find the roots without a graph, factorise the quadratic into two brackets, then set each bracket equal to zero and solve, since the whole expression equals zero whenever either bracket does.
y = x² - 5x + 6 factorises to (x - 2)(x - 3), so the roots are x = 2 and x = 3.
Finding the turning point by completing the square
Writing a quadratic in the form (x + a)² + b reveals the turning point directly: it sits at (-a, b), since the squared bracket is smallest, equal to zero, when x = -a.
y = x² - 6x + 5 completes the square to (x - 3)² - 4, so the turning point is at (3, -4).
Worked Examples
Three exam-style questions, fully solved.
For y = x² + 2, work out the value of y when x = -2, x = 0 and x = 3.
Easy- 1.Substitute x = -2: y = (-2)² + 2 = 6
- 2.Substitute x = 0: y = 0² + 2 = 2
- 3.Substitute x = 3: y = 3² + 2 = 11
Answer: y = 6, y = 2, y = 11
By factorising, find the roots of y = x² - 5x + 6.
Medium- 1.Find two numbers that multiply to give 6 and add to give -5: -2 and -3
- 2.Write the factorised form: (x - 2)(x - 3) = 0
- 3.Set each bracket to zero: x - 2 = 0 or x - 3 = 0
Answer: x = 2 and x = 3
By completing the square, find the turning point of y = x² - 6x + 5.
Hard- 1.Halve the coefficient of x to get the number inside the bracket: -6 ÷ 2 = -3, giving (x - 3)²
- 2.Subtract the square of that number to correct the constant: (x - 3)² - 9 + 5 = (x - 3)² - 4
- 3.Read the turning point from the completed square form: (3, -4)
Answer: (3, -4)
Avoid These
The most common mistakes students make.
Squaring a negative x value incorrectly, for example treating (-2)² as -4 instead of the correct positive value, 4.
Confusing the roots, where the curve crosses the x-axis, with the y-intercept, where the curve crosses the y-axis.
When factorising, choosing two numbers that multiply to give the constant term but do not add to give the correct coefficient of x.
When completing the square, forgetting to subtract the square of the halved coefficient after adding it inside the bracket, leaving the wrong constant term.
Reading the turning point's x-coordinate with the wrong sign, for example giving (a, b) instead of (-a, b) from the completed square form (x + a)² + b.
FAQ
Questions parents and students ask.
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