Chapter 2: Quadratic Functions
2.1 Definition and Standard Form
Definition:
A quadratic function is a function of the form:
\[
f(x) = ax^2 + bx + c \quad \text{where } a, b, c \in \mathbb{R} \text{ and } a \neq 0
\]
Shape and direction:
- If \(a > 0\): parabola opens upwards (U-shape) → has a minimum value
- If \(a < 0\): parabola opens downwards (\(\cap\)-shape) → has a maximum value
Proof: The sign of \(a\) determines direction
Consider \(f(x) = ax^2 + bx + c\). As \(x \to \pm\infty\):
- If \(a > 0\): \(ax^2 \to +\infty\), so \(f(x) \to +\infty\) (graph goes up on both ends)
- If \(a < 0\): \(ax^2 \to -\infty\), so \(f(x) \to -\infty\) (graph goes down on both ends)
The parabola cannot go up on one side and down on the other because the \(x^2\) term dominates for large \(|x|\).
2.2 Completing the Square
Purpose:
Completing the square rewrites \(ax^2 + bx + c\) in the form:
\[
f(x) = a(x - h)^2 + k
\]
where \((h, k)\) is the vertex (turning point) of the parabola.
Derivation:
\[
\begin{aligned}
f(x) &= ax^2 + bx + c \\
&= a\left(x^2 + \frac{b}{a}x\right) + c \\
&= a\left(x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right) + c \\
&= a\left[\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2\right] + c \\
&= a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)
\end{aligned}
\]
\(ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right)\)
So the vertex is at:
\[
h = -\frac{b}{2a}, \quad k = c - \frac{b^2}{4a}
\]
Proof: The vertex is the maximum or minimum point
For \(a > 0\): \((x - h)^2 \ge 0\), so \(a(x - h)^2 \ge 0\). Hence \(f(x) \ge k\). The minimum value \(k\) occurs when \((x - h)^2 = 0\), i.e., \(x = h\).
For \(a < 0\): \(a(x - h)^2 \le 0\), so \(f(x) \le k\). The maximum value \(k\) occurs at \(x = h\).
Worked Example 2.1 (Completing the square – non-calculator)
Express \(f(x) = 2x^2 - 12x + 7\) in the form \(a(x - h)^2 + k\).
Solution:
\[
\begin{aligned}
f(x) &= 2x^2 - 12x + 7 \\
&= 2(x^2 - 6x) + 7 \\
&= 2[(x - 3)^2 - 9] + 7 \\
&= 2(x - 3)^2 - 18 + 7 \\
&= 2(x - 3)^2 - 11
\end{aligned}
\]
Vertex: \((3, -11)\). Since \(a = 2 > 0\), this is a minimum point.
Mauritian context – maximum profit
A shop in Port Louis sells sugar. The daily profit \(P(x)\) in rupees is given by:
\[
P(x) = -5x^2 + 100x - 300
\]
where \(x\) is the price per kg in rupees. Find the price that maximises profit, and the maximum profit.
Solution:
\[
\begin{aligned}
P(x) &= -5x^2 + 100x - 300 \\
&= -5(x^2 - 20x) - 300 \\
&= -5[(x - 10)^2 - 100] - 300 \\
&= -5(x - 10)^2 + 500 - 300 \\
&= -5(x - 10)^2 + 200
\end{aligned}
\]
Vertex at \(x = 10\), maximum value \(200\).
Answer: Price = Rs 10 per kg gives maximum profit = Rs 200.
2.3 Maximum and Minimum Values
From completed square form \(f(x) = a(x - h)^2 + k\):
| Condition |
Shape |
Turning point |
Value |
| \(a > 0\) |
U-shape |
Minimum |
\(k\) at \(x = h\) |
| \(a < 0\) |
\(\cap\)-shape |
Maximum |
\(k\) at \(x = h\) |
Finding the range for a given domain:
If the domain is restricted to \([p, q]\), the range is found by:
- Evaluate \(f(p)\) and \(f(q)\)
- If the vertex \(h\) lies inside \([p, q]\), include \(k\) as the min or max
- The range is between the smallest and largest of these values
Worked Example 2.3 (Range on a restricted domain)
\(f(x) = x^2 - 4x + 5\) for \(1 \le x \le 4\). Find the range.
Solution:
Complete the square: \(x^2 - 4x + 5 = (x - 2)^2 + 1\). Vertex at \(x = 2\).
Evaluate:
- \(f(1) = 1 - 4 + 5 = 2\)
- \(f(2) = 1\) (minimum)
- \(f(4) = 16 - 16 + 5 = 5\)
Smallest value = 1, largest = 5.
Answer: Range: \([1, 5]\)
2.4 The Discriminant and Nature of Roots
The quadratic equation \(ax^2 + bx + c = 0\) has solutions:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
The expression under the square root is the discriminant:
\[
\Delta = b^2 - 4ac
\]
Nature of roots:
| \(\Delta\) |
Nature of roots |
Graphical meaning |
| \(\Delta > 0\) |
Two distinct real roots |
Curve cuts x-axis twice |
| \(\Delta = 0\) |
One repeated real root |
Curve touches x-axis (tangent) |
| \(\Delta < 0\) |
No real roots |
Curve does not meet x-axis |
Proof: Why \(\Delta = 0\) gives a repeated root
If \(\Delta = 0\), then:
\[
x = \frac{-b \pm 0}{2a} = -\frac{b}{2a}
\]
Both the \(+\) and \(-\) give the same value. Hence one root (repeated).
Worked Example 2.4 (Using the discriminant)
Find the values of \(k\) for which the equation \(x^2 + kx + 9 = 0\) has two distinct real roots.
Solution:
\(\Delta = k^2 - 4(1)(9) = k^2 - 36\)
For two distinct real roots: \(\Delta > 0\)
\[
k^2 - 36 > 0 \Rightarrow k^2 > 36 \Rightarrow k < -6 \text{ or } k > 6
\]
Answer: \(k < -6\) or \(k > 6\)
2.5 Quadratic Inequalities
Steps to solve \(ax^2 + bx + c > 0\) (or \(\ge, <, \le\)):
- Solve \(ax^2 + bx + c = 0\) to find the boundary points (roots)
- Sketch the parabola (or use a sign table)
- Read the solution set from the sketch
Key fact:
For \(a > 0\), the quadratic is positive outside the roots and negative between the roots.
Proof: Sign pattern for \(a > 0\)
If \(r_1 < r_2\) are the roots, then \(ax^2 + bx + c = a(x - r_1)(x - r_2)\).
For \(x > r_2\): both factors positive \(\rightarrow\) product positive
For \(r_1 < x < r_2\): \((x - r_1) > 0\), \((x - r_2) < 0\) \(\rightarrow\) product negative
For \(x < r_1\): both factors negative \(\rightarrow\) product positive
Multiply by \(a > 0\) preserves signs.
Worked Example 2.5 (Quadratic inequality – non-calculator)
Solve \(x^2 - 5x + 6 \le 0\).
Solution:
Factorise: \((x - 2)(x - 3) \le 0\)
Roots at \(x = 2\) and \(x = 3\). Since \(a = 1 > 0\), parabola opens upwards.
It is below (or on) the x-axis between the roots.
Answer: \(2 \le x \le 3\)
2.6 Line–Curve Intersection
Given a line \(y = mx + k\) and a curve \(y = ax^2 + bx + c\):
- Equate: \(ax^2 + bx + c = mx + k\)
- Rearrange to form a quadratic: \(ax^2 + (b - m)x + (c - k) = 0\)
- The discriminant \(\Delta\) of this quadratic determines intersection:
| \(\Delta\) |
Intersection type |
| \(\Delta > 0\) |
Two distinct points (secant) |
| \(\Delta = 0\) |
One point (tangent) |
| \(\Delta < 0\) |
No intersection |
Proof: \(\Delta = 0\) gives tangency
When the line and curve meet at exactly one point, the quadratic has a repeated root. A repeated root means the line touches the curve without crossing – this is the definition of a tangent.
Worked Example 2.6 (Tangent condition – non-calculator)
Find the value of \(k\) such that the line \(y = 4x + k\) is a tangent to the curve \(y = x^2 + 2x + 3\).
Solution:
Equate: \(x^2 + 2x + 3 = 4x + k\)
\[
x^2 + 2x + 3 - 4x - k = 0
\]
\[
x^2 - 2x + (3 - k) = 0
\]
For tangency, \(\Delta = 0\):
\[
(-2)^2 - 4(1)(3 - k) = 0
\]
\[
4 - 12 + 4k = 0
\]
\[
-8 + 4k = 0 \Rightarrow k = 2
\]
Answer: \(k = 2\)
2.7 Sketching Quadratic Graphs
To sketch \(y = ax^2 + bx + c\):
- Direction: \(a > 0\) (U-shape), \(a < 0\) (\(\cap\)-shape)
- Vertex: \(h = -\frac{b}{2a}\), \(k = f(h)\) (or from completed square)
- y-intercept: \((0, c)\)
- x-intercepts (roots): Solve \(ax^2 + bx + c = 0\) (if \(\Delta \ge 0\))
- Axis of symmetry: vertical line \(x = h\)
Worked Example 2.7 (Sketching)
Sketch \(y = -x^2 + 4x + 5\). Label vertex and intercepts.
Solution:
\(a = -1 < 0\) \(\rightarrow\) \(\cap\)-shape.
Complete square:
\[
-x^2 + 4x + 5 = -(x^2 - 4x) + 5 = -[(x - 2)^2 - 4] + 5 = -(x - 2)^2 + 9
\]
Vertex: \((2, 9)\) (maximum)
y-intercept: \((0, 5)\)
x-intercepts: \(-x^2 + 4x + 5 = 0 \Rightarrow x^2 - 4x - 5 = 0 \Rightarrow (x - 5)(x + 1) = 0\)
Roots: \(x = 5\) and \(x = -1\) \(\rightarrow\) points \((5, 0)\) and \((-1, 0)\)
Axis of symmetry: \(x = 2\)
Chapter 2 Summary
| Concept |
Key points |
| Standard form |
\(f(x) = ax^2 + bx + c\), \(a \neq 0\) |
| Completed square |
\(a(x - h)^2 + k\), vertex \((h, k)\) |
| Max/min |
\(a > 0\): minimum at vertex; \(a < 0\): maximum |
| Discriminant |
\(\Delta = b^2 - 4ac\) |
| Nature of roots |
\(\Delta > 0\): two real; \(\Delta = 0\): one repeated; \(\Delta < 0\): none |
| Line–curve intersection |
Substitute, form quadratic, use \(\Delta\) |
Exercises – Chapter 2
Easy (Drill – Non-Calculator)
- For each quadratic, state whether the graph has a maximum or minimum, and find the vertex:
(a) \(f(x) = x^2 - 6x + 10\)
(b) \(f(x) = -2x^2 + 8x - 5\)
(c) \(f(x) = 3x^2 + 12x + 7\)
- Find the discriminant and state the nature of the roots for:
(a) \(2x^2 + 5x - 3 = 0\)
(b) \(x^2 - 4x + 4 = 0\)
(c) \(3x^2 + x + 2 = 0\)
- Solve the inequalities:
(a) \(x^2 - 4x + 3 > 0\)
(b) \(x^2 + 2x - 8 \le 0\)
Medium (Examination Style)
- Express \(f(x) = 2x^2 - 8x + 5\) in completed square form. Hence find the minimum value of \(f(x)\) and the value of \(x\) at which it occurs.
- Mauritian context: A farmer in Flacq has a rectangular field. The length is \(x\) metres and the width is \((20 - x)\) metres.
(a) Show that the area \(A = 20x - x^2\).
(b) Find the maximum possible area and the dimensions that give it.
(c) If the area must be at least 75 m², find the possible range of \(x\).
- Find the values of \(k\) for which the equation \(x^2 + kx + (k + 3) = 0\) has two distinct real roots.
- The line \(y = 2x + c\) is a tangent to the curve \(y = x^2 - 3x + 1\). Find the value of \(c\).
Hard (Challenge for A*)
- The quadratic function \(f(x) = x^2 - 4kx + 5k^2 + 2k + 3\), where \(k\) is a constant.
(a) Show that \(f(x) \ge 2\) for all real values of \(x\).
(b) Find the value of \(k\) for which the minimum value of \(f(x)\) is exactly 3.
- The line \(y = mx + 3\) intersects the curve \(y = x^2 - 2x + 5\) at two distinct points.
(a) Show that \(m^2 + 4m - 12 < 0\).
(b) Hence find the range of possible values of \(m\).
- Mauritian context: A mobile network company models its monthly profit \(P\) (in thousands of rupees) as:
\[
P(x) = -4x^2 + 48x - 80
\]
where \(x\) is the number of thousands of subscribers.
(a) Find the number of subscribers that maximises profit.
(b) Find the maximum profit.
(c) The company breaks even when \(P(x) = 0\). Find the two break-even points.
(d) For what range of subscribers does the company make a profit (\(P(x) > 0\))?