This note covers the quadratic function \(f(x) = ax^2 + bx + c\) and its three useful forms: standard, factorised and vertex form. Each form shows a different feature of the parabola, so choosing the right one makes graphs and optimisation problems much quicker.
Key ideas
A quadratic function has the highest power of \(x\) equal to 2. Its graph is a smooth curve called a parabola. It is symmetrical about a vertical line, the axis of symmetry, and the point where it turns is the vertex.
- If \(a \gt 0\), the parabola opens upwards and the vertex is a minimum point.
- If \(a \lt 0\), the parabola opens downwards and the vertex is a maximum point.
- A larger value of \(|a|\) makes the parabola narrower.
Standard form: \(f(x) = ax^2 + bx + c\). The \(y\)-intercept is \(c\), because \(f(0) = c\). The axis of symmetry is \(x = -\dfrac{b}{2a}\). Substitute this \(x\)-value back into \(f\) to find the \(y\)-coordinate of the vertex.
Factorised form: \(f(x) = a(x - p)(x - q)\). The \(x\)-intercepts are at \(x = p\) and \(x = q\). By symmetry, the axis of symmetry is halfway between them: \(x = \dfrac{p + q}{2}\). This form is best when you know the zeros.
Vertex form: \(f(x) = a(x - h)^2 + k\). The vertex is at \((h, k)\) and the axis of symmetry is \(x = h\). This form is best when you know the turning point, or when you want the greatest or least value.
Completing the square turns standard form into vertex form. Take \(f(x) = ax^2 + bx + c\):
- Factor out \(a\) from the \(x^2\) and \(x\) terms.
- Inside the bracket, add and subtract the square of half the \(x\) coefficient.
- Write the first three terms as a perfect square, and collect the constants.
Sketching. To sketch a quadratic, decide whether it opens up or down, then mark the \(y\)-intercept, the \(x\)-intercepts if there are any, and the vertex. The intercepts and the vertex give the shape. If the vertex form shows that the vertex is above the \(x\)-axis and the parabola opens up, there are no \(x\)-intercepts.
Optimisation. Many real problems ask for the greatest area, the greatest profit or the least cost. If the situation is modelled by a quadratic, the answer is at the vertex. Write the model, find the vertex, and then interpret both coordinates in context.
Finding the \(x\)-intercepts (solving \(f(x) = 0\)) is covered more fully in the next note on quadratic equations.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(f(x) = ax^2 + bx + c\) | Standard form. \(y\)-intercept is \(c\). |
| \(x = -\dfrac{b}{2a}\) | Axis of symmetry from the standard form. |
| \(f(x) = a(x - p)(x - q)\) | Factorised form. Zeros at \(p\) and \(q\). |
| \(f(x) = a(x - h)^2 + k\) | Vertex form. Vertex \((h, k)\), axis \(x = h\). |
| \(x = \dfrac{p + q}{2}\) | Axis of symmetry when the zeros are \(p\) and \(q\). |
| Completing the square | Method to move from standard form to vertex form. |
| Vertex | The maximum point if \(a \lt 0\); the minimum point if \(a \gt 0\). |
Worked examples
Example 1 Write \(f(x) = 2x^2 - 12x + 7\) in the form \(a(x - h)^2 + k\). Hence state the vertex, the axis of symmetry and the \(x\)-intercepts.
Step 1: factor out 2 from the first two terms. \(f(x) = 2(x^2 - 6x) + 7\).
Step 2: half of \(-6\) is \(-3\), and \((-3)^2 = 9\). Inside the bracket write \(x^2 - 6x = (x - 3)^2 - 9\).
Step 3: substitute back. \(f(x) = 2[(x - 3)^2 - 9] + 7 = 2(x - 3)^2 - 18 + 7 = 2(x - 3)^2 - 11\).
Step 4: read the features. The vertex is \((3, -11)\), and the axis of symmetry is \(x = 3\). Since \(a = 2 \gt 0\), it is a minimum point.
Step 5: the \(x\)-intercepts. Solve \(2(x - 3)^2 - 11 = 0\), so \((x - 3)^2 = 5.5\) and \(x - 3 = \pm\sqrt{5.5}\). Then \(x = 3 \pm 2.345\), so \(x = 0.655\) or \(x = 5.35\) (to 3 s.f.).
Answer: \(f(x) = 2(x - 3)^2 - 11\); vertex \((3, -11)\); axis \(x = 3\); \(x\)-intercepts \(0.655\) and \(5.35\).
Example 2 A parabola crosses the \(x\)-axis at \(x = -1\) and \(x = 5\) and passes through the point \((2, 18)\). Find its equation in factorised form and in standard form, and state its vertex.
Step 1: use factorised form. \(y = a(x + 1)(x - 5)\).
Step 2: substitute the point \((2, 18)\). \(18 = a(3)(-3) = -9a\), so \(a = -2\).
Step 3: factorised form is \(y = -2(x + 1)(x - 5)\).
Step 4: expand. \((x + 1)(x - 5) = x^2 - 4x - 5\). Multiply by \(-2\): \(y = -2x^2 + 8x + 10\).
Step 5: vertex. The axis is halfway between the zeros: \(x = \dfrac{-1 + 5}{2} = 2\). At \(x = 2\), \(y = 18\). So the vertex is \((2, 18)\), a maximum because \(a \lt 0\).
Answer: \(y = -2(x + 1)(x - 5) = -2x^2 + 8x + 10\); vertex \((2, 18)\).
Example 3 A farmer has 60 m of fencing and builds a rectangular pen against a straight wall. The wall forms one side, so the fencing is used for the other three sides. Let the two sides at right angles to the wall each have length \(x\) metres. Find the greatest possible area of the pen.
Step 1: the side parallel to the wall is \(60 - 2x\) metres, because two sides of length \(x\) use up \(2x\) metres.
Step 2: area \(A = x(60 - 2x) = -2x^2 + 60x\). The pen needs positive lengths, so \(0 \lt x \lt 30\).
Step 3: the coefficient of \(x^2\) is negative, so the vertex is a maximum. The axis is \(x = -\dfrac{60}{2(-2)} = 15\), which lies in the allowed range.
Step 4: when \(x = 15\), \(A = 15 \times (60 - 30) = 15 \times 30 = 450\).
Answer: the greatest area is 450 m², when the sides at right angles to the wall are 15 m each and the side along the wall is 30 m.
Common errors
- Common error: getting the sign of \(h\) wrong in vertex form. The bracket \((x - 3)^2\) gives \(h = 3\), but \((x + 3)^2\) gives \(h = -3\). The value of \(k\) keeps its own sign.
- Common error: forgetting to multiply the added constant by \(a\) when completing the square. In \(2[(x - 3)^2 - 9]\), the \(-9\) becomes \(-18\).
- Common error: using the factorised form with \(a = 1\) when a point shows that \(a\) is not 1. Always find \(a\) using the extra point.
- Common error: giving the \(x\)-coordinate only for the vertex when the question asks for the greatest value, or the other way round.
- Common error: forgetting to check that the \(x\)-value of the vertex is allowed in the context.
Exam tip: choose the form that gives the answer directly. Use factorised form for intercepts, vertex form for the turning point, and standard form for the \(y\)-intercept.
Check yourself
- Write \(x^2 + 6x + 1\) in the form \((x + p)^2 + q\) and state the vertex.
Answer
Half of 6 is 3, so \(x^2 + 6x = (x + 3)^2 - 9\). Then \(x^2 + 6x + 1 = (x + 3)^2 - 8\). The vertex is \((-3, -8)\). - Find the equation of the axis of symmetry of \(y = 3x^2 - 12x + 5\).
Answer
\(x = -\dfrac{b}{2a} = -\dfrac{-12}{2(3)} = 2\). The axis is \(x = 2\). - A parabola has its vertex at \((1, -3)\) and passes through \((3, 5)\). Find its equation in vertex form.
Answer
Write \(y = a(x - 1)^2 - 3\). Substitute \((3, 5)\): \(5 = a(2)^2 - 3\), so \(4a = 8\) and \(a = 2\). The equation is \(y = 2(x - 1)^2 - 3\). - Find the least value of \(x^2 - 8x + 20\) and the value of \(x\) at which it occurs.
Answer
\(x^2 - 8x + 20 = (x - 4)^2 + 4\). The least value is 4, when \(x = 4\).