This note covers quadratic and cubic functions used as models: how to read their graphs, what the parameters mean in a real context, and how to use your GDC to find maximum values, roots and solutions. Many applied questions, from the path of a ball to the profit of a business, rest on these two families of functions.
Key ideas
A model is a function that describes a real situation. Quadratic models describe situations where a quantity rises and then falls (or falls and then rises) smoothly. Cubic models describe situations with up to two turning points, such as volume formed by cutting and folding a sheet.
Quadratic models. The general form is \(f(x) = ax^2 + bx + c\) with \(a \neq 0\). The graph is a parabola.
- If \(a \gt 0\) the parabola opens upwards and has a minimum point. If \(a \lt 0\) it opens downwards and has a maximum point. In a model with a maximum, such as a thrown ball, expect \(a\) to be negative.
- The axis of symmetry is the vertical line \(x = -\dfrac{b}{2a}\). The turning point (vertex) lies on this line, so its \(x\)-coordinate is \(-\dfrac{b}{2a}\). Substitute this back into the function to get the \(y\)-coordinate.
- The \(y\)-intercept is \(c\). In a model it is the starting value, for example the height at time \(t = 0\).
- The roots (zeros) are the \(x\)-values where \(f(x) = 0\). Find them with the quadratic formula, by factorising, or by using the GDC graph or equation solver.
- Other forms are useful: factorised form \(f(x) = a(x - p)(x - q)\) shows the roots \(p\) and \(q\), and vertex form \(f(x) = a(x - h)^2 + k\) shows the vertex \((h, k)\).
Cubic models. The general form is \(f(x) = ax^3 + bx^2 + cx + d\) with \(a \neq 0\). A cubic graph has one, two or three real roots. It has either two turning points (a local maximum and a local minimum) or none. For large positive \(x\), the graph goes up if \(a \gt 0\) and down if \(a \lt 0\). Turning points, roots and intersections are found using the GDC, not by hand.
Using the GDC. Draw the graph in a suitable window. Then use the "maximum" or "minimum" feature to find turning points, the "zero" or "root" feature to find roots, and the "intersect" feature to solve \(f(x) = k\). You can also use the polynomial equation solver for \(ax^3 + bx^2 + cx + d = 0\).
Domain and context. A model is only valid for sensible inputs. Time cannot be negative, a length cannot be zero or negative, and a box cannot have a side longer than the sheet it comes from. Always check whether each solution fits the domain, and discard those that do not.
Fitting a model through points. A quadratic has three unknowns, so three points on the curve give three equations in \(a\), \(b\) and \(c\). Write the equations, then solve them with the GDC simultaneous equation solver or by substitution. Regression on the GDC, covered in the note on evaluating models, can also fit a quadratic or cubic to a data set.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(f(x) = ax^2 + bx + c\) | General quadratic. \(c\) is the \(y\)-intercept. The sign of \(a\) decides whether there is a maximum or a minimum. |
| \(x = -\dfrac{b}{2a}\) | Axis of symmetry and \(x\)-coordinate of the vertex. |
| \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) | Roots of a quadratic. Real roots exist only when \(b^2 - 4ac \geq 0\). |
| \(f(x) = a(x - p)(x - q)\) | Factorised form. The roots are \(x = p\) and \(x = q\). |
| \(f(x) = a(x - h)^2 + k\) | Vertex form. The vertex is \((h, k)\). |
| \(f(x) = ax^3 + bx^2 + cx + d\) | General cubic. \(d\) is the \(y\)-intercept. Up to three roots and up to two turning points. |
| Local maximum / minimum | A turning point where the graph changes from rising to falling (or falling to rising). Found with the GDC. |
| Domain of a model | The set of input values that make sense in the context. |
Worked examples
Example 1 A stone is thrown upwards from a footbridge. Its height above the river, in metres, after \(t\) seconds is modelled by \(h(t) = -4.9t^2 + 19.6t + 1.5\), for \(t \geq 0\). Find (a) the height of the bridge, (b) the maximum height and when it occurs, (c) when the stone hits the river.
(a) The height at \(t = 0\) is the \(y\)-intercept: \(h(0) = 1.5\). The bridge is 1.5 m above the river.
(b) The turning point is at \(t = -\dfrac{19.6}{2 \times (-4.9)} = \dfrac{-19.6}{-9.8} = 2\) seconds.
Then \(h(2) = -4.9(4) + 19.6(2) + 1.5 = -19.6 + 39.2 + 1.5 = 21.1\) m.
(c) The stone reaches the river when \(h(t) = 0\). Use the zero feature on your GDC, or the quadratic formula. The roots are \(t = 4.075\ldots\) and \(t = -0.0751\ldots\)
The negative value is outside the domain \(t \geq 0\), so it is rejected.
Answer: (a) 1.5 m; (b) 21.1 m at \(t = 2\) s; (c) \(t = 4.08\) s.
Example 2 A small company sells phone cases at a price of \(x\) euros. Its weekly profit, in euros, is modelled by \(P(x) = -2x^2 + 60x - 300\). Find the price that gives the greatest profit, the greatest profit, and the range of prices for which the company makes a profit of at least €100.
The coefficient of \(x^2\) is negative, so \(P\) has a maximum.
The vertex is at \(x = -\dfrac{60}{2 \times (-2)} = 15\).
\(P(15) = -2(225) + 900 - 300 = -450 + 900 - 300 = 150\).
For a profit of at least €100 we need \(-2x^2 + 60x - 300 \geq 100\), which is \(-2x^2 + 60x - 400 \geq 0\).
Divide by \(-2\) and reverse the inequality: \(x^2 - 30x + 200 \leq 0\). This factorises as \((x - 10)(x - 20) \leq 0\).
The graph of \(P\) is above the line \(P = 100\) between the two intersection points, so \(10 \leq x \leq 20\).
Answer: price €15 gives a maximum profit of €150; a profit of at least €100 needs a price from €10 to €20.
Example 3 A rectangular sheet of card measures 30 cm by 20 cm. A square of side \(x\) cm is cut from each corner and the sides are folded up to make an open box. (a) Show that the volume is \(V = 4x^3 - 100x^2 + 600x\) and state the domain. (b) Find the maximum volume. (c) Find the values of \(x\) for which \(V = 500\).
(a) The base has length \(30 - 2x\) and width \(20 - 2x\). The height is \(x\).
\(V = x(30 - 2x)(20 - 2x) = x(600 - 100x + 4x^2) = 4x^3 - 100x^2 + 600x\).
The width must be positive, so \(20 - 2x \gt 0\) and \(x \lt 10\). Also \(x \gt 0\). The domain is \(0 \lt x \lt 10\).
(b) Graph \(V\) on the GDC with \(0 \lt x \lt 10\) and use the maximum feature. The maximum is at \(x = 3.92\ldots\), where \(V = 1056.3\ldots\) cm³.
(c) Solve \(4x^3 - 100x^2 + 600x - 500 = 0\) using the polynomial solver, or find where \(V\) meets the line \(y = 500\). The roots are \(x = 0.990\ldots\), \(7.77\ldots\) and \(16.2\ldots\)
The value 16.2 is outside the domain, so it is rejected.
Answer: (b) maximum volume 1060 cm³ (3 s.f.); (c) \(x = 0.990\) or \(x = 7.77\).
Common errors
- Giving the \(x\)-coordinate of the vertex when the question asks for the maximum value. The maximum value is the \(y\)-coordinate.
- Leaving in a root that is outside the domain, such as a negative time or a side length longer than the sheet.
- Forgetting to reverse an inequality sign when dividing both sides by a negative number.
- Using \(x = \dfrac{b}{2a}\) instead of \(x = -\dfrac{b}{2a}\) for the axis of symmetry.
- Rounding too early. Keep full GDC values in later steps and round only the final answer.
Exam tip: When a question gives a domain, write it down before you use the GDC and set the graph window to match. State clearly which GDC solutions you keep and which you reject, with a reason.
Check yourself
- Find the minimum value of \(f(x) = 3x^2 - 12x + 5\).
Answer
The vertex is at \(x = -\dfrac{-12}{2 \times 3} = 2\). Then \(f(2) = 3(4) - 24 + 5 = -7\). The minimum value is \(-7\). - A farmer uses 40 m of fencing to enclose a rectangular pen against a long wall, so only three sides need fencing. The two sides touching the wall have length \(x\) m. Write the area as a function of \(x\) and find its maximum.
Answer
The side parallel to the wall is \(40 - 2x\), so \(A = x(40 - 2x) = -2x^2 + 40x\). The vertex is at \(x = -\dfrac{40}{2 \times (-2)} = 10\). \(A(10) = -200 + 400 = 200\). The maximum area is 200 m², for \(0 \lt x \lt 20\). - The cubic \(f(x) = x^3 - 6x^2 + 9x + 2\) is drawn on a GDC. Which two turning points does it have?
Answer
Using the maximum and minimum features, the local maximum is at \((1, 6)\) and the local minimum is at \((3, 2)\). Check: \(f(1) = 1 - 6 + 9 + 2 = 6\) and \(f(3) = 27 - 54 + 27 + 2 = 2\). - A quadratic model \(y = ax^2 + bx + c\) passes through \((0, 3)\), \((1, 8)\) and \((2, 11)\). Find the model and predict \(y\) when \(x = 4\).
Answer
From \((0, 3)\), \(c = 3\). From \((1, 8)\): \(a + b + 3 = 8\), so \(a + b = 5\). From \((2, 11)\): \(4a + 2b + 3 = 11\), so \(2a + b = 4\). Subtracting gives \(a = -1\) and then \(b = 6\). The model is \(y = -x^2 + 6x + 3\). At \(x = 4\): \(y = -16 + 24 + 3 = 11\).