Parabolas By Completing The Square
Theory
Completing the square rewrites \(y = x^2 + bx + c\) as \(y = (x - h)^2 + k\), so the turning point \((h, k)\) can be read straight off. This Year 10 maths page shows how to complete the square to sketch a parabola with its intercepts and range, to find exact \(x\)-intercepts with surds, to factorise a quadratic that has no whole-number factors, and to tell how many \(x\)-intercepts a parabola has.
A perfect square \((x + h)^2 = x^2 + 2hx + h^2\) has a constant equal to the square of half the coefficient of \(x\). So \(x^2 + bx\) needs \(\left(\dfrac{b}{2}\right)^2\) added to become a perfect square: that is completing the square.
To keep the expression equal, add and subtract the same number: \(x^2 + bx + c = \left(x + \dfrac{b}{2}\right)^2 - \left(\dfrac{b}{2}\right)^2 + c\).
If the leftover number is negative, as in \((x - 1)^2 - 6\), write \(6 = (\sqrt{6})^2\) and use the difference of two squares. If it is positive, as in \((x - 1)^2 + 6\), the expression is always positive and has no real linear factors.
The form \(a(x + h)^2 + k\) is called the completed-square form. It also shows the smallest (or largest) value of the quadratic, \(k\), when \(x = -h\).
For a parabola, completing the square on \(y = x^2 + bx + c\) gives the turning-point form \(y = (x - h)^2 + k\). The turning point is \((h, k)\): the parabola can then be sketched without a table of values.
Turning-point form of a parabola and its \(x\)-intercepts:
How to sketch a parabola by completing the square
- Prepare: if the \(x^2\) coefficient is not \(1\), take it out of the \(x^2\) and \(x\) terms. If it is negative, take out \(-1\).
- Complete the square inside the bracket: add and subtract the square of half the coefficient of \(x\). Whatever you subtract inside the bracket is multiplied by \(a\) when you remove it.
- Read the turning point \((h, k)\) from \(y = a(x - h)^2 + k\), and the concavity from \(a\).
- \(y\)-intercept: \(y = c\) from the original equation.
- \(x\)-intercepts: set \(y = 0\), make \((x - h)^2\) the subject and take \(\pm\) square roots (exact answers may be surds). A negative value means there are none.
- Sketch a smooth curve, symmetric about \(x = h\), through the labelled points, and state the range \(y \ge k\) or \(y \le k\).
How to factorise by completing the square
- Prepare: if the \(x^2\) coefficient is not \(1\), take it out of the \(x\) terms (or of everything). If it is negative, take out \(-1\).
- Halve the coefficient of \(x\), square it, and add and subtract that number.
- Write the perfect square and collect the constants: \((x + h)^2 - d^2\).
- Factorise as a difference of two squares, write the number as the square of a surd if needed, and simplify each bracket.
- Check by expanding, or by substituting a value of \(x\).
Using the completed square
- Turning point given: write \(y = (x - h)^2 + k\), expand, and match it with \(y = x^2 + bx + c\) to find \(b\) and \(c\). With one more point, use \(y = a(x - h)^2 + k\) and solve for \(a\).
- Families of parabolas: with a letter such as \(m\) in the equation, complete the square to find the turning point in terms of \(m\), then apply the condition given (passes through a point, lies on a line) (Example 4).
- Finding unknowns: if the factors are \((x + h + d)(x + h - d)\), they differ by \(2d\). A given difference of factors, or a given perfect square, fixes \(d\) and hence the constant.
- No real factors: if the completed square is \((x + h)^2 + k\) with \(k > 0\), the expression is always positive, so it has no real linear factors (Example 4).
(i) \(x^2 + 10x + 21\)
(ii) \(x^2 - 6x + 2\), with surds in simplest form.
(i) Half of \(10\) is \(5\), and \(5^2 = 25\). Add and subtract \(25\), then use the difference of two squares.
| \(x^2 + 10x + 21\) | \(=\) | \(x^2 + 10x + 25 - 25 + 21\) |
| \(\) | \(=\) | \((x + 5)^2 - 4\) |
| \(\) | \(=\) | \((x + 5)^2 - 2^2\) |
| \(\) | \(=\) | \((x + 5 - 2)(x + 5 + 2)\) |
| \(\) | \(=\) | \(\boldsymbol{(x + 3)(x + 7)}\) |
(ii) Half of \(-6\) is \(-3\), and \((-3)^2 = 9\). The leftover number \(7\) is not a perfect square, so write it as \((\sqrt{7})^2\).
| \(x^2 - 6x + 2\) | \(=\) | \((x - 3)^2 - 9 + 2\) |
| \(\) | \(=\) | \((x - 3)^2 - (\sqrt{7})^2\) |
| \(\) | \(=\) | \(\boldsymbol{(x - 3 - \sqrt{7})(x - 3 + \sqrt{7})}\) |
(ii) Find the exact \(x\)-intercepts of \(y = x^2 + 6x + 4\).
(i) Half of \(-2\) is \(-1\); add and subtract \(1\).
| \(y\) | \(=\) | \(x^2 - 2x + 1 - 1 - 8\) |
| \(\) | \(=\) | \(\boldsymbol{(x - 1)^2 - 9}\) |
Turning point \((1, -9)\), a minimum (\(a = 1 \gt 0\)). \(y\)-intercept \(-8\). Range \(\boldsymbol{y \ge -9}\).
| \((x - 1)^2\) | \(=\) | \(9\) |
| \(x - 1\) | \(=\) | \(\pm 3\) |
| \(x\) | \(=\) | \(-2 \text{ or } 4\) |
(ii) Complete the square, then set \(y = 0\):
| \(y\) | \(=\) | \((x + 3)^2 - 9 + 4 = (x + 3)^2 - 5\) |
| \((x + 3)^2\) | \(=\) | \(5\) |
| \(x\) | \(=\) | \(\boldsymbol{-3 \pm \sqrt{5}}\) |
(i) \(y = -x^2 + 6x - 5\)
(ii) \(y = 3x^2 - 12x + 7\)
(i) Take out \(-1\) from the \(x\) terms; the sign of the \(x\) term flips inside the bracket.
| \(y\) | \(=\) | \(-(x^2 - 6x) - 5\) |
| \(\) | \(=\) | \(-(x^2 - 6x + 9 - 9) - 5\) |
| \(\) | \(=\) | \(-(x - 3)^2 + 9 - 5\) |
| \(\) | \(=\) | \(\boldsymbol{-(x - 3)^2 + 4}\) |
Maximum turning point \((3, 4)\). \(y\)-intercept \(-5\). \((x - 3)^2 = 4\) gives \(x = 1\) or \(5\).
(ii) Take out \(3\) from the \(x\) terms only. The \(-4\) inside the bracket becomes \(-12\) outside.
| \(y\) | \(=\) | \(3(x^2 - 4x) + 7\) |
| \(\) | \(=\) | \(3\left[(x - 2)^2 - 4\right] + 7\) |
| \(\) | \(=\) | \(\boldsymbol{3(x - 2)^2 - 5}\) |
Turning point \((2, -5)\), a minimum. It is below the \(x\)-axis and the parabola is concave up, so there are two \(x\)-intercepts.
(ii) For each value of \(t\), \(y = x^2 - 2tx + t^2 - 2t\). Show that the turning point is \((t, -2t)\), and find the values of \(t\) for which the parabola passes through \((0, 3)\).
(i) Use \(y = a(x + 2)^2 - 5\) and substitute \((0, -3)\):
| \(-3\) | \(=\) | \(4a - 5\) |
| \(a\) | \(=\) | \(\dfrac{1}{2}\) |
So \(y = \boldsymbol{\dfrac{1}{2}(x + 2)^2 - 5}\). Let \(y = 0\):
| \((x + 2)^2\) | \(=\) | \(10\) |
| \(x\) | \(=\) | \(\boldsymbol{-2 \pm \sqrt{10}}\) |
(ii) Complete the square with \(t\) in place: half of \(-2t\) is \(-t\).
| \(y\) | \(=\) | \((x^2 - 2tx + t^2) - 2t = (x - t)^2 - 2t\) |
So the turning point is \((t, -2t)\). Substitute \((0, 3)\):
| \(t^2 - 2t\) | \(=\) | \(3\) |
| \((t - 3)(t + 1)\) | \(=\) | \(0\) |
| \(t\) | \(=\) | \(\boldsymbol{3} \text{ or } \boldsymbol{-1}\) |
The parabolas are \(y = (x - 3)^2 - 6\) and \(y = (x + 1)^2 + 2\). Every turning point \((t, -2t)\) lies on the line \(y = -2x\).
Common pitfalls
Frequently asked questions
How do you complete the square?
Take half of the coefficient of \(x\) and square it. Add and subtract that number, then write the first three terms as a squared bracket: \(x^2 + 8x + 3 = (x + 4)^2 - 16 + 3 = (x + 4)^2 - 13\).
Why do you add and subtract the same number?
Adding a number makes a perfect square, and subtracting the same number keeps the expression equal to the original. Together they add zero.
How do you factorise a quadratic with surds?
Complete the square to get \((x + h)^2 - d\). Write \(d\) as \((\sqrt{d})^2\), then use \(a^2 - b^2 = (a - b)(a + b)\).
How do you complete the square when the coefficient of x squared is not 1?
Take the coefficient out of the x squared and x terms, complete the square inside the bracket, then multiply the subtracted square by that coefficient when you remove the bracket.
How can you tell a quadratic has no real factors?
Complete the square. If you get a bracket squared plus a positive number, the expression is always positive, so it is never zero and has no real linear factors.