Chap1. Theory of quadratic functions and quadratic equations(8h)

# Chap1. Theory of quadratic functions and quadratic equations(8h) ## Objectifs et prérequis This chapter aims to provide a comprehensive understanding of quadratic functions and equations, enabling students to analyze, transform, solve, and interpret quadratic expressions in various forms. By the end of this chapter, students will be able to: identify and manipulate quadratic expressions; apply simple transformations to quadratic graphs; determine the nature of roots using the discriminant; solve absolute value equations involving quadratic expressions; complete the square; use the quadratic formula; factorize quadratics; relate roots to coefficients; exploit symmetric properties; analyze properties and graphs of quadratic functions; and interpret scaling transformations. Prerequisites: Proficiency in algebraic manipulation, solving linear equations, understanding functions and their notation, basic graphing skills on the Cartesian plane, and familiarity with real numbers and inequalities. --- ## I. Quadratic expressions and equations A quadratic expression is a polynomial of degree two in one variable, generally written in the form $ ax^2 + bx + c $, where $ a, b, c \in \mathbb{R} $ and $ a \neq 0 $. A quadratic equation is formed when such an expression is set equal to zero: $ ax^2 + bx + c = 0 $. > **Definition**: A quadratic expression in $ x $ is any expression of the form $ ax^2 + bx + c $ with $ a \neq 0 $, $ a, b, c \in \mathbb{R} $. **Properties**: - The graph of $ y = ax^2 + bx + c $ is a parabola. - If $ a > 0 $, the parabola opens upwards; if $ a < 0 $, it opens downwards. - The expression can be rewritten in vertex form via completing the square. **Example 1**: Identify whether $ 3x^2 - 5x + 2 $ is quadratic. Solution: Yes, because it is of degree 2 and the coefficient of $ x^2 $ is 3 ($ \neq 0 $). **Example 2**: Is $ 4x + 7 $ quadratic? Solution: No, because the highest power of $ x $ is 1. **Exercise 1**: Which of the following are quadratic expressions? (a) $ x^2 + 4x - 5 $ (b) $ 2x^3 + x^2 $ (c) $ -x^2 + 3 $ (d) $ 5 $ (e) $ (x+1)(x-2) $ **Correction 1**: (a) Yes – degree 2, $ a=1 \neq 0 $ (b) No – degree 3 (c) Yes – $ a=-1 \neq 0 $ (d) No – degree 0 (e) Yes – expands to $ x^2 - x - 2 $, degree 2 --- ## II. Simple transformation of the form $ y = f(x \pm a) $, $ a \in \mathbb{R} $ This transformation shifts the graph of $ y = f(x) $ horizontally. Replacing $ x $ with $ x - a $ shifts the graph $ a $ units to the right; replacing $ x $ with $ x + a $ shifts it $ a $ units to the left. > **Rule**: For $ y = f(x - a) $, graph shifts right by $ a $ units; for $ y = f(x + a) $, graph shifts left by $ a $ units. **Example 1**: Given $ f(x) = x^2 $, sketch $ y = f(x - 3) = (x - 3)^2 $. Solution: The graph of $ y = x^2 $ is shifted 3 units to the right. Vertex moves from $ (0,0) $ to $ (3,0) $. **Example 2**: Sketch $ y = f(x + 2) = (x + 2)^2 $ for $ f(x) = x^2 $. Solution: Shift left by 2 units. Vertex at $ (-2, 0) $. **Exercise 2**: Let $ f(x) = x^2 - 4x + 3 $. (a) Write the expression for $ y = f(x - 1) $ and simplify. (b) Describe the transformation from $ y = f(x) $ to $ y = f(x + 2) $. **Correction 2**: (a) $ f(x - 1) = (x - 1)^2 - 4(x - 1) + 3 = x^2 - 2x + 1 - 4x + 4 + 3 = x^2 - 6x + 8 $ (b) Replacing $ x $ with $ x + 2 $ shifts the graph 2 units to the left. --- ## III. Simple transformation of the form $ y = f(x) \pm a $, $ a \in \mathbb{R} $ This transformation shifts the graph vertically. Adding $ a $ shifts the graph up by $ a $ units; subtracting $ a $ shifts it down by $ a $ units. > **Rule**: For $ y = f(x) + a $, graph shifts up by $ a $ units; for $ y = f(x) - a $, graph shifts down by $ a $ units. **Example 1**: For $ f(x) = x^2 $, graph $ y = f(x) + 4 = x^2 + 4 $. Solution: Shift up by 4 units. Vertex moves from $ (0,0) $ to $ (0,4) $. **Example 2**: Graph $ y = f(x) - 3 = x^2 - 3 $. Solution: Shift down by 3 units. Vertex at $ (0,-3) $. **Exercise 3**: Let $ f(x) = 2x^2 + 3x - 1 $. (a) Find $ y = f(x) + 5 $ and describe the transformation. (b) What is the vertex of $ y = f(x) - 2 $ if the vertex of $ f(x) $ is at $ \left(-\frac{3}{4}, -\frac{17}{8}\right) $? **Correction 3**: (a) $ f(x) + 5 = 2x^2 + 3x + 4 $; shift up by 5 units. (b) Vertex shifts down by 2: $ \left(-\frac{3}{4}, -\frac{17}{8} - 2\right) = \left(-\frac{3}{4}, -\frac{33}{8}\right) $ --- ## IV. Application of nature of roots The nature of roots of a quadratic equation $ ax^2 + bx + c = 0 $ is determined by the discriminant $ \Delta = b^2 - 4ac $. > **Rule**: - If $ \Delta > 0 $: two distinct real roots - If $ \Delta = 0 $: one real root (repeated) - If $ \Delta < 0 $: no real roots (two complex conjugate roots) **Example 1**: Determine the nature of roots for $ 2x^2 - 4x + 2 = 0 $. Solution: $ a=2, b=-4, c=2 $; $ \Delta = (-4)^2 - 4(2)(2) = 16 - 16 = 0 $ → one real root (repeated). **Example 2**: For $ x^2 + x + 1 = 0 $, $ \Delta = 1 - 4 = -3 < 0 $ → no real roots. **Exercise 4**: Find the values of $ k $ for which the equation $ x^2 + kx + 9 = 0 $ has: (a) two distinct real roots (b) equal roots (c) no real roots **Correction 4**: $ \Delta = k^2 - 36 $ (a) $ k^2 - 36 > 0 $ → $ k < -6 $ or $ k > 6 $ (b) $ k^2 - 36 = 0 $ → $ k = \pm 6 $ (c) $ k^2 - 36 < 0 $ → $ -6 < k < 6 $ --- ## V. Absolute value equations of the form $ |f(x)| = a $ To solve $ |f(x)| = a $, where $ a \geq 0 $, we solve the two equations: $ f(x) = a $ and $ f(x) = -a $. If $ a < 0 $, there is no solution. > **Rule**: $ |f(x)| = a $ ($ a \geq 0 $) is equivalent to $ f(x) = a $ or $ f(x) = -a $. **Example 1**: Solve $ |x^2 - 4| = 5 $. Solution: Case 1: $ x^2 - 4 = 5 $ → $ x^2 = 9 $ → $ x = \pm 3 $ Case 2: $ x^2 - 4 = -5 $ → $ x^2 = -1 $ → no real solution Solutions: $ x = -3, 3 $ **Example 2**: Solve $ |2x^2 - 8| = 0 $. Solution: $ |f(x)| = 0 $ implies $ f(x) = 0 $ → $ 2x^2 - 8 = 0 $ → $ x^2 = 4 $ → $ x = \pm 2 $ **Exercise 5**: Solve: (a) $ |x^2 - 5x + 6| = 2 $ (b) $ |3x^2 - 12| = 9 $ **Correction 5**: (a) Case 1: $ x^2 - 5x + 6 = 2 $ → $ x^2 - 5x + 4 = 0 $ → $ (x-1)(x-4)=0 $ → $ x=1,4 $ Case 2: $ x^2 - 5x + 6 = -2 $ → $ x^2 - 5x + 8 = 0 $ → $ \Delta = 25 - 32 = -7 < 0 $ → no real Solutions: $ x = 1, 4 $ (b) Case 1: $ 3x^2 - 12 = 9 $ → $ 3x^2 = 21 $ → $ x^2 = 7 $ → $ x = \pm \sqrt{7} $ Case 2: $ 3x^2 - 12 = -9 $ → $ 3x^2 = 3 $ → $ x^2 = 1 $ → $ x = \pm 1 $ Solutions: $ x = -\sqrt{7}, -1, 1, \sqrt{7} $ --- ## VI. Method of completing the square This method rewrites a quadratic expression in the form $ a(x - h)^2 + k $, revealing the vertex $ (h, k) $. > **Rule**: To complete the square for $ ax^2 + bx + c $: 1. Factor out $ a $ from $ x^2 $ and $ x $ terms (if $ a \neq 1 $) 2. Add and subtract $ \left(\frac{b}{2a}\right)^2 $ inside the parentheses 3. Simplify to vertex form **Example 1**: Complete the square for $ x^2 + 6x + 5 $. Solution: $ x^2 + 6x + 5 = (x^2 + 6x + 9) - 9 + 5 = (x+3)^2 - 4 $ **Example 2**: Complete the square for $ 2x^2 - 8x + 3 $. Solution: Factor 2: $ 2(x^2 - 4x) + 3 $ Complete inside: $ x^2 - 4x = (x-2)^2 - 4 $ So: $ 2[(x-2)^2 - 4] + 3 = 2(x-2)^2 - 8 + 3 = 2(x-2)^2 - 5 $ **Exercise 6**: Complete the square for: (a) $ x^2 - 10x + 21 $ (b) $ -3x^2 + 12x - 7 $ **Correction 6**: (a) $ x^2 - 10x + 21 = (x^2 - 10x + 25) - 25 + 21 = (x-5)^2 - 4 $ (b) Factor -3: $ -3(x^2 - 4x) - 7 $ $ x^2 - 4x = (x-2)^2 - 4 $ → $ -3[(x-2)^2 - 4] - 7 = -3(x-2)^2 + 12 - 7 = -3(x-2)^2 + 5 $ --- ## VII. Quadratic Formula method The quadratic formula provides the roots of $ ax^2 + bx + c = 0 $ directly: > **Formula**: $ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ \Delta = b^2 - 4ac $ **Example 1**: Solve $ 2x^2 + 3x - 2 = 0 $ using the quadratic formula. Solution: $ a=2, b=3, c=-2 $ $ \Delta = 9 - 4(2)(-2) = 9 + 16 = 25 $ $ x = \dfrac{-3 \pm \sqrt{25}}{4} = \dfrac{-3 \pm 5}{4} $ → $ x = \dfrac{2}{4} = \frac{1}{2} $ or $ x = \dfrac{-8}{4} = -2 $ **Example 2**: Solve $ x^2 - 4x + 5 = 0 $. Solution: $ \Delta = 16 - 20 = -4 $ $ x = \dfrac{4 \pm \sqrt{-4}}{2} = \dfrac{4 \pm 2i}{2} = 2 \pm i $ **Exercise 7**: Solve using the quadratic formula: (a) $ 3x^2 - 5x + 2 = 0 $ (b) $ x^2 + 6x + 10 = 0 $ **Correction 7**: (a) $ a=3, b=-5, c=2 $; $ \Delta = 25 - 24 = 1 $ $ x = \dfrac{5 \pm 1}{6} $ → $ x = 1 $ or $ x = \frac{4}{6} = \frac{2}{3} $ (b) $ a=1, b=6, c=10 $; $ \Delta = 36 - 40 = -4 $ $ x = \dfrac{-6 \pm \sqrt{-4}}{2} = -3 \pm i $ --- ## VIII. Solving by factorization This method expresses the quadratic as a product of two linear factors and uses the zero-product property. > **Rule**: If $ ax^2 + bx + c = (px + q)(rx + s) $, then solutions are $ x = -\frac{q}{p} $ and $ x = -\frac{s}{r} $. **Example 1**: Solve $ x^2 - 5x + 6 = 0 $ by factorization. Solution: Find two numbers that multiply to 6 and add to -5: -2 and -3 $ (x-2)(x-3) = 0 $ → $ x = 2 $ or $ x = 3 $ **Example 2**: Solve $ 2x^2 + 7x + 3 = 0 $. Solution: Look for factors of $ 2 \times 3 = 6 $ that add to 7: 6 and 1 Rewrite: $ 2x^2 + 6x + x + 3 = 0 $ Factor: $ 2x(x+3) + 1(x+3) = (2x+1)(x+3) = 0 $ → $ x = -\frac{1}{2} $ or $ x = -3 $ **Exercise 8**: Solve by factorization: (a) $ x^2 + x - 12 = 0 $ (b) $ 3x^2 - 10x + 3 = 0 $ **Correction 8**: (a) Factors of -12 that add to 1: 4 and -3 $ (x+4)(x-3) = 0 $ → $ x = -4 $ or $ x = 3 $ (b) Factors of $ 3 \times 3 = 9 $ that add to -10: -9 and -1 $ 3x^2 - 9x - x + 3 = 3x(x-3) -1(x-3) = (3x-1)(x-3) = 0 $ → $ x = \frac{1}{3} $ or $ x = 3 $ --- ## IX. Relation between roots and coefficients of quadratic equation For a quadratic equation $ ax^2 + bx + c = 0 $ with roots $ \alpha $ and $ \beta $: > **Formulas**: - Sum of roots: $ \alpha + \beta = -\dfrac{b}{a} $ - Product of roots: $ \alpha \beta = \dfrac{c}{a} $ **Example 1**: If the roots of $ 2x^2 - 7x + 3 = 0 $ are $ \alpha $ and $ \beta $, find $ \alpha + \beta $ and $ \alpha \beta $. Solution: $ \alpha + \beta = -\dfrac{-7}{2} = \frac{7}{2} $ $ \alpha \beta = \dfrac{3}{2} $ **Example 2**: Find a quadratic equation with roots $ 3 $ and $ -2 $. Solution: Sum = $ 3 + (-2) = 1 $, Product = $ 3 \times (-2) = -6 $ Equation: $ x^2 - (\text{sum})x + \text{product} = 0 $ → $ x^2 - x - 6 = 0 $ **Exercise 9**: (a) If the roots of $ x^2 - 5x + k = 0 $ are $ \alpha $ and $ \beta $, and $ \alpha + \beta = 5 $, find $ k $ given that $ \alpha \beta = 6 $. (b) Form the quadratic equation whose roots are $ \frac{1}{2} $ and $ -3 $. **Correction 9**: (a) From sum: $ -\frac{-5}{1} = 5 $ (consistent). Product: $ \alpha \beta = k = 6 $ → $ k = 6 $ (b) Sum = $ \frac{1}{2} + (-3) = -\frac{5}{2} $, Product = $ \frac{1}{2} \times (-3) = -\frac{3}{2} $ Equation: $ x^2 - (-\frac{5}{2})x + (-\frac{3}{2}) = 0 $ → $ x^2 + \frac{5}{2}x - \frac{3}{2} = 0 $ Multiply by 2: $ 2x^2 + 5x - 3 = 0 $ --- ## X. Symmetric properties of quadratic equations Symmetric expressions in the roots $ \alpha $ and $ \beta $ can be expressed in terms of $ \alpha + \beta $ and $ \alpha \beta $. > **Rule**: Any symmetric function of $ \alpha $ and $ \beta $ can be written using $ S = \alpha + \beta $ and $ P = \alpha \beta $. **Common symmetric forms**: - $ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = S^2 - 2P $ - $ \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = S^3 - 3SP $ - $ \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{S}{P} $ (if $ P \neq 0 $) - $ \alpha - \beta = \pm \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \pm \sqrt{S^2 - 4P} $ **Example 1**: If $ \alpha $ and $ \beta $ are roots of $ x^2 - 4x + 3 = 0 $, find $ \alpha^2 + \beta^2 $. Solution: $ S = 4, P = 3 $ $ \alpha^2 + \beta^2 = 4^2 - 2 \times 3 = 16 - 6 = 10 $ **Example 2**: For the same equation, find $ \dfrac{1}{\alpha} + \dfrac{1}{\beta} $. Solution: $ \dfrac{S}{P} = \dfrac{4}{3} $ **Exercise 10**: Let $ \alpha $ and $ \beta $ be the roots of $ 2x^2 - 6x + 5 = 0 $. Find: (a) $ \alpha^2 + \beta^2 $ (b) $ \alpha^3 + \beta^3 $ (c) $ \dfrac{1}{\alpha} + \dfrac{1}{\beta} $ **Correction 10**: $ S = -\frac{-6}{2} = 3 $, $ P = \frac{5}{2} $ (a) $ \alpha^2 + \beta^2 = S^2 - 2P = 9 - 2 \times \frac{5}{2} = 9 - 5 = 4 $ (b) $ \alpha^3 + \beta^3 = S^3 - 3SP = 27 - 3 \times 3 \times \frac{5}{2} = 27 - \frac{45}{2} = \frac{54 - 45}{2} = \frac{9}{2} $ (c) $ \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{S}{P} = \dfrac{3}{5/2} = \dfrac{6}{5} $ --- ## XI. Properties of quadratic Functions A quadratic function $ f(x) = ax^2 + bx + c $ has the following properties: > **Properties**: - Domain: $ \mathbb{R} $ - Range: - If $ a > 0 $: $ [f(x_{\text{min}}), \infty) $ - If $ a < 0 $: $ (-\infty, f(x_{\text{max}})] $ - Vertex: $ \left( -\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right) \right) $ - Axis of symmetry: $ x = -\dfrac{b}{2a} $ - y-intercept: $ (0, c) $ - x-intercepts: depend on discriminant (0, 1, or 2 real roots) **Example 1**: For $ f(x) = -x^2 + 4x - 3 $, find the vertex, axis of symmetry, and range. Solution: $ a = -1 < 0 $ → opens downwards Vertex: $ x = -\dfrac{4}{2(-1)} = 2 $, $ f(2) = -4 + 8 - 3 = 1 $ → vertex $ (2,1) $ Axis: $ x = 2 $ Range: $ (-\infty, 1] $ **Exercise 11**: For $ f(x) = 3x^2 - 6x + 2 $, determine: (a) vertex (b) axis of symmetry (c) range (d) y-intercept (e) number of x-intercepts **Correction 11**: (a) $ x = -\frac{-6}{2 \cdot 3} = 1 $, $ f(1) = 3 - 6 + 2 = -1 $ → vertex $ (1, -1) $ (b) Axis: $ x = 1 $ (c) $ a = 3 > 0 $ → range $ [-1, \infty) $ (d) y-intercept: $ f(0) = 2 $ → $ (0, 2) $ (e) $ \Delta = (-6)^2 - 4 \cdot 3 \cdot 2 = 36 - 24 = 12 > 0 $ → two distinct real roots → two x-intercepts --- ## XII. Graphs of quadratic functions The graph of $ y = ax^2 + bx + c $ is a parabola. Key features for sketching: - Direction (based on sign of $ a $) - Vertex (turning point) - Axis of symmetry - Intercepts with axes - Symmetry about the axis **Example 1**: Sketch $ y = x^2 - 4x + 3 $. Solution: Factor: $ (x-1)(x-3) $ → x-intercepts at $ x=1, 3 $ Vertex: $ x = 2 $, $ y = 4 - 8 + 3 = -1 $ → $ (2, -1) $ y-intercept: $ x=0 $ → $ y=3 $ → $ (0,3) $ Since $ a=1>0 $, opens upwards. Plot points and draw symmetric parabola. **Exercise 12**: Sketch the graph of $ y = -2x^2 + 8x - 6 $, labeling: - vertex - axis of symmetry - y-intercept - x-intercepts (if any) **Correction 12**: Factor: $ -2(x^2 - 4x + 3) = -2(x-1)(x-3) $ → x-intercepts at $ x=1, 3 $ Vertex: $ x = -\frac{8}{2(-2)} = 2 $, $ y = -2(4) + 16 - 6 = -8 + 16 - 6 = 2 $ → vertex $ (2, 2) $ Axis: $ x = 2 $ y-intercept: $ x=0 $ → $ y = -6 $ → $ (0, -6) $ Since $ a = -2 < 0 $, opens downwards. Plot: points $ (0,-6) $, $ (1,0) $, $ (2,2) $, $ (3,0) $, symmetric about $ x=2 $. --- ## XIII. Simple transformation of the form $ f(x) = a x^2 $, $ a \in \mathbb{R} $ This transformation scales the graph of $ y = x^2 $ vertically. If $ |a| > 1 $, it stretches; if $ 0 < |a| < 1 $, it compresses; if $ a < 0 $, it reflects over the x-axis. > **Rule**: For $ y = a x^2 $: - If $ a > 1 $: vertical stretch - If $ 0 < a < 1 $: vertical compression - If $ a = 1 $: identical to $ y = x^2 $ - If $ a < 0 $: reflection over x-axis, plus stretch/compression based on $ |a| $ **Example 1**: Compare $ y = 2x^2 $ and $ y = x^2 $. Solution: For each $ x $, $ y $ is doubled → vertical stretch by factor 2. Vertex remains $ (0,0) $, opens upwards. **Example 2**: Graph $ y = -\frac{1}{2}x^2 $. Solution: Reflection over x-axis (due to negative sign), and vertical compression by factor $ \frac{1}{2} $. Opens downwards, wider than $ y = -x^2 $. **Exercise 13**: Describe the transformation from $ y = x^2 $ to: (a) $ y = -3x^2 $ (b) $ y = \frac{1}{4}x^2 $ (c) $ y = -0.5x^2 $ **Correction 13**: (a) Reflection over x-axis and vertical stretch by factor 3 (b) Vertical compression by factor $ \frac{1}{4} $ (no reflection) (c) Reflection over x-axis and vertical compression by factor $ 0.5 $ --- ## Synthèse du chapitre - Quadratic expressions are degree-2 polynomials; equations set them to zero. - Horizontal shifts: $ y = f(x \pm a) $ shifts graph left/right. - Vertical shifts: $ y = f(x) \pm a $ shifts graph up/down. - Nature of roots determined by discriminant $ \Delta = b^2 - 4ac $. - Absolute value equations $ |f(x)| = a $ split into $ f(x) = \pm a $ (for $ a \geq 0 $). - Completing the square converts to vertex form $ a(x-h)^2 + k $. - Quadratic formula: $ x = \frac{-b \pm \sqrt{\Delta}}{2a} $. - Factorization uses zero-product property after expressing as product of linear factors. - Sum and product of roots: $ \alpha + \beta = -b/a $, $ \alpha \beta = c/a $. - Symmetric root expressions reducible to sum and product. - Quadratic functions have parabolic graphs with vertex, axis, direction, and intercepts. - Scaling: $ y = a x^2 $ stretches/compresses and may reflect the base parabola. --- ## Glossaire - **Quadratic expression**: A polynomial of degree two in one variable, $ ax^2 + bx + c $, $ a \neq 0 $. - **Discriminant**: $ \Delta = b^2 - 4ac $, determines nature of roots of $ ax^2 + bx + c = 0 $. - **Vertex**: The turning point of a parabola, $ \left( -\frac{b}{2a}, f\left(-\frac{b}{2a}\right) \right) $. - **Axis of symmetry**: The vertical line $ x = -\frac{b}{2a} $ about which the parabola is symmetric. - **Completing the square**: Rewriting $ ax^2 + bx + c $ as $ a(x-h)^2 + k $. - **Quadratic formula**: $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, gives roots of quadratic equation. - **Factorization**: Expressing quadratic as product of two linear factors to solve via zero-product property. - **Sum of roots**: For $ ax^2 + bx + c = 0 $, $ \alpha + \beta = -\frac{b}{a} $. - **Product of roots**: $ \alpha \beta = \frac{c}{a} $. - **Absolute value equation**: Equation of form $ |f(x)| = a $, solved by considering $ f(x) = \pm a $. - **Transformation**: Change in position, size, or orientation of a graph (e.g., shift, stretch, reflection). - **Parabola**: The U-shaped graph of a quadratic function.