Quadratic Functions: Sketching, Maximum and Minimum: PECTAA Class 11 Maths, Punjab Board

Throw a ball upward and its height first rises and then falls, tracing a parabola with one highest point. Every quadratic function has such a turning point. Unit 3 of the book shows how to find it by sketching and by completing the square.

Unit 3 of the PECTAA Mathematics 11 book is Theory of Quadratic Functions. Section 3.1 covers sketching a quadratic function (3.1.1) and finding its maximum or minimum value by completing the square (3.1.2). This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Quadratic function (section 3.1). A quadratic function is a polynomial function of degree two, written in standard form

f(x) = ax² + bx + c, where a, b, c are real and a ≠ 0.

Shape (section 3.1.1). The graph is a parabola. It opens upward when a > 0 and downward when a < 0.

• Vertex (h, k): the tip of the parabola, a turning point where the graph changes direction.
• Axis of symmetry: the vertical line x = h through the vertex.
• a > 0: the vertex is a minimum point.
• a < 0: the vertex is a maximum point.

Sketching and analysing. To sketch, find:

1. the y-intercept (put x = 0);
2. the x-intercepts (solve ax² + bx + c = 0);
3. the vertex, with h = −b/(2a) and k = f(h).

To analyse, say whether the vertex is a maximum or a minimum, and give the intervals where the function increases and decreases.

Completing the square (section 3.1.2). This rewrites f(x) = ax² + bx + c in vertex form

f(x) = a(x − h)² + k, where h = −b/(2a) and k = c − b²/(4a).

• If a > 0, the minimum value of f is k, at x = h.
• If a < 0, the maximum value of f is k, at x = h.

The reason is simple: (x − h)² is never negative and is zero only at x = h.

Key points

• The sign of a decides everything: up and minimum, or down and maximum.
• h = −b/(2a) always gives the x-coordinate of the vertex.
• For a < 0, the function increases to the left of h and decreases to the right; for a > 0 it is the reverse.
• The maximum or minimum value is k, the y-coordinate of the vertex.

Worked examples

Example 1 (book Example 1). Sketch and analyse y = −x² − 2x + 3.

1. y-intercept: y = 3 at x = 0.
2. x-intercepts: −x² − 2x + 3 = 0, so x² + 2x − 3 = 0, (x + 3)(x − 1) = 0, giving x = −3 and x = 1.
3. Vertex: h = −(−2)/(2 × (−1)) = −1, and k = −(−1)² − 2(−1) + 3 = −1 + 2 + 3 = 4. Vertex (−1, 4).
4. a = −1 < 0, so (−1, 4) is a maximum point.
5. The function increases on ]−∞, −1[ and decreases on ]−1, ∞[.

Example 2 (book Example 2). Find the maximum or minimum of f(x) = −2x² + 4x + 3 by completing the square.

1. f(x) = −2(x² − 2x) + 3.
2. Add and subtract 1 inside: −2(x² − 2x + 1 − 1) + 3 = −2[(x − 1)² − 1] + 3.
3. f(x) = −2(x − 1)² + 2 + 3 = −2(x − 1)² + 5.
4. a = −2 < 0, so the maximum value is 5, at x = 1.

Example 3 (book Example 3). Find the maximum or minimum of f(x) = x² − 2x − 3.

1. a = 1, b = −2, c = −3.
2. h = −(−2)/(2 × 1) = 1.
3. k = c − b²/(4a) = −3 − 4/4 = −4.
4. a = 1 > 0, so the minimum value is −4, at x = 1.

Common mistakes in the exam

1. Sign error in h. h = −b/(2a); with b = −2 and a = −1, h = −1, not +1.
2. Squaring a negative carelessly. In −x², x = −1 gives −(−1)² = −1, not +1.
3. Calling a vertex a minimum when a < 0. Check the sign of a first.
4. Forgetting to multiply back when completing the square: −2 × (−1) = +2.
5. Giving the x-value as the maximum. The maximum value is k, attained at x = h.
6. Mixing interval ends. The function changes direction at x = h.

Exam technique

• Lay out sketch questions as: intercepts → vertex → nature of vertex → increasing and decreasing intervals.
• In completing-the-square questions, show the "add and subtract" step clearly.
• Check k by substituting h back into the original function.
• Check your board's notice for the sections in your paper.

Practice questions

1. Does y = 3x² − 6x + 1 have a maximum or a minimum? Answer: A minimum, because a = 3 > 0.
2. Find the vertex of y = x² + 4x + 1. Answer: h = −2, k = 4 − 8 + 1 = −3. Vertex (−2, −3).
3. Write y = x² − 6x + 5 in vertex form. Answer: y = (x − 3)² − 4.
4. Find the maximum value of y = −x² + 4x. Answer: y = −(x − 2)² + 4, so the maximum is 4 at x = 2.
5. Where does y = x² − 2x − 3 cross the x-axis? Answer: (x − 3)(x + 1) = 0, so at x = 3 and x = −1.

Ready to practise this? Continue in 1st Year Mathematics.