Quadratic Functions: Forms, Characteristics, and Graphing

Overview of Quadratic Functions

  • Real-world applications modeled by curved trajectories:

    • The path of a ball thrown into the air.

    • The trajectory of a basketball moving toward a hoop.

    • The shape of water spurting out of a fountain.

  • Comparison between linear and quadratic functions:

    • Linear Function: Produces a straight-line graph because the highest exponent of xx is 1 (e.g., y=x+2y = x + 2).

    • Quadratic Function: Produces a curved graph called a parabola because the highest exponent of xx is 2 (e.g., y=x2+2y = x^2 + 2).

  • Key mathematical applications: Used to study curved motion paths, distance, height, maximum values, and minimum values.

  • Definition: A quadratic function is a function that can be written in the form f(x)=ax2+bx+cf(x) = ax^2 + bx + c, where aa, bb, and cc are real numbers, a≠0a \neq 0, and the highest exponent of xx is 2.

  • Graph: The shape formed by graphing a quadratic function is called a parabola.

Identifying Quadratic Functions

  • Classification based on the highest exponent of the variable xx:

    • Highest Exponent 1: Linear function

    • Highest Exponent 2: Quadratic function

    • Highest Exponent 3: Cubic function

  • Worked Examples:

    • Example 1: Determine whether f(x)=3x2+5x−2f(x) = 3x^2 + 5x - 2 is quadratic.

    • Evaluation: The highest exponent of xx is 2 and a=3≠0a = 3 \neq 0.

    • Conclusion: The function is quadratic with coefficients a=3a = 3, b=5b = 5, and c = -2$.\n * Example 2: Determine whether f(x) = -4x^2 + 7 is quadratic.\n * Evaluation: The highest exponent of xis2andis 2 anda = -4 eq 0.\n * Conclusion: The function is quadratic with coefficients a = -4,,b = 0,and, andc = 7$.

  • Guided Practice 1 - Function Identification:

    1. f(x)=6x2−3x+4f(x) = 6x^2 - 3x + 4 (Quadratic)

    2. f(x)=−x2+5x−6f(x) = -x^2 + 5x - 6 (Quadratic)

    3. f(x)=2x+8f(x) = 2x + 8 (Not quadratic - Linear)

    4. f(x)=(x−5)2−4f(x) = (x - 5)^2 - 4 (Quadratic)

    5. f(x)=x3+4f(x) = x^3 + 4 (Not quadratic - Cubic)

    6. f(x)=9−2x2f(x) = 9 - 2x^2 (Quadratic)

    7. f(x)=−2x2+10xf(x) = -2x^2 + 10x (Quadratic)

    8. f(x)=(x−2)(x−5)f(x) = (x - 2)(x - 5) (Quadratic)

Three Forms of Quadratic Functions

  • A quadratic function can be expressed in three distinct algebraic forms: Standard Form, Vertex Form, and Intercept Form.

  • 1. Standard Form

    • Equation: f(x)=ax2+bx+cf(x) = ax^2 + bx + c

    • Structural parameters: aa is the coefficient of x2x^2, bb is the coefficient of xx, and cc is the constant term.

    • Direct information extracted: Coefficients aa, bb, cc, and the yy-intercept directly at (0,c)(0, c).

    • Worked Examples:

    • Example 1: Given f(x)=2x2−8x+6f(x) = 2x^2 - 8x + 6, identify aa, bb, and cc.

      • Comparing with standard form yields a=2a = 2, b=−8b = -8, c = 6$.\n * Example 2: Given f(x) = -2(x + 2)^2 + 5,converttostandardformandidentify, convert to standard form and identifya,,b,and, andc.\n * Step 1 (Expand the squared binomial): f(x) = -2(x^2 + 4x + 4) + 5\n * Step 2 (Distribute -2):):f(x) = -2x^2 - 8x - 8 + 5\n * Step 3 (Simplify): f(x) = -2x^2 - 8x - 3\n * Result: a = -2,,b = -8,,c = -3\n * Example 3: Given f(x) = 2(x - 4)(x + 3),converttostandardformandidentify, convert to standard form and identifya,,b,and, andc.\n * Step 1 (Expand the two binomials): (x - 4)(x + 3) = x^2 + 3x - 4x - 12 = x^2 - x - 12\n * Step 2 (Multiply by 2): f(x) = 2(x^2 - x - 12) = 2x^2 - 2x - 24\n * Result: a = 2,,b = -2,,c = -24\n * Guided Practice 2 - Identify a,,b,and, andc:\n 1. f(x) = x^2 + 6x + 5\n 2. f(x) = 3x^2 - 4x - 7\n 3. f(x) = (x - 2)^2 - 1\n 4. f(x) = (x - 3)^2 - 4\n 5. f(x) = 2(x - 3)(x + 1)\n 6. f(x) = (x - 1)(x - 3)\n\n* **2. Vertex Form**\n * Equation: f(x) = a(x - h)^2 + k\n * Structural parameters: adeterminesthedirectionandwidthoftheparabola,determines the direction and width of the parabola,xistheinputvariable,is the input variable,histheis thex−coordinateofthevertex,and-coordinate of the vertex, andkistheis they-coordinate of the vertex.\n * Sign convention for h: Always use the opposite sign of the value inside the parentheses:\n * (x - 4) \rightarrow h = 4\n * (x + 4) \rightarrow h = -4\n * Worked Examples:\n * Example 1: Given f(x) = 2(x - 3)^2 + 4,identify, identifya,,h,,k, and the vertex.\n * Comparing with vertex form: a = 2,,h = 3,,k = 4\n * Vertex: (h, k) = (3, 4)\n * Example 2: Given f(x) = -x^2 + 6x + 7, convert to vertex form and find the vertex.\n * Step 1 (Group x−terms):-terms):f(x) = (-x^2 + 6x) + 7 = -(x^2 - 6x) + 7\n * Step 2 (Half of x−coefficientsquared):Coefficientis-coefficient squared): Coefficient is-6;halfis; half is-6 \div 2 = -3;squareit:; square it:(-3)^2 = 9\n * Step 3 (Add and subtract 9 inside parentheses): f(x) = -(x^2 - 6x + 9 - 9) + 7\n * Step 4 (Factor perfect square trinomial): f(x) = -[(x - 3)^2 - 9] + 7\n * Step 5 (Distribute negative sign): f(x) = -(x - 3)^2 + 9 + 7\n * Step 6 (Simplify): f(x) = -(x - 3)^2 + 16\n * Vertex Form: f(x) = -(x - 3)^2 + 16\n * Parameters: a = -1,,h = 3,,k = 16\n * Vertex: (3, 16)\n * Guided Practice 3 - Identify Vertex:\n 1. f(x) = (x - 2)^2 + 3\n 2. f(x) = 2(x + 1)^2 - 4\n 3. f(x) = x^2 - 4x + 3\n 4. f(x) = x^2 - 6x + 8\n 5. f(x) = (x - 2)(x - 4)\n 6. f(x) = (x + 1)(x - 3)\n\n* **3. Intercept Form**\n * Equation: f(x) = a(x - p)(x - q)\n * Structural parameters: adeterminesopeningdirectionandwidth,determines opening direction and width,prepresentsthefirstrepresents the firstx−intercept,and-intercept, andqrepresentsthesecondrepresents the secondx-intercept.\n * x−interceptscoordinates:-intercepts coordinates:(p, 0)andand(q, 0).\n * Worked Examples:\n * Example 1: Given f(x) = 2(x - 3)(x + 1),findthe, find thex-intercepts.\n * Step 1 (Set function equal to zero): 2(x - 3)(x + 1) = 0\n * Step 2 (Factor/isolate product): (x - 3)(x + 1) = 0\n * Step 3 (Set each factor to zero): x - 3 = 0 \implies x = 3;;x + 1 = 0 \implies x = -1\n * Result: x−interceptsare-intercepts are(3, 0)andand(-1, 0).\n * Example 2: Given f(x) = x^2 - 5x + 6,findthe, find thex-intercepts.\n * Step 1 (Set function equal to zero): x^2 - 5x + 6 = 0\n * Step 2 (Factor the quadratic): (x - 2)(x - 3) = 0\n * Step 3 (Set each factor to zero): x - 2 = 0 \implies x = 2;;x - 3 = 0 \implies x = 3\n * Result: x−interceptsare-intercepts are(2, 0)andand(3, 0).\n * Guided Practice 4 - Determine x-intercepts:\n 1. f(x) = (x - 2)(x - 5)\n 2. f(x) = x^2 + 2x - 3\n 3. f(x) = x^2 - 2x - 3\n 4. f(x) = 2(x - 1)(x + 2)\n 5. f(x) = -(x + 5)(x - 2)\n\n# Characteristics of Quadratic Functions\n\n* **1. Direction of Opening**\n * Governed by parameter aininf(x) = ax^2 + bx + c:\n * If a > 0: Parabola opens **upward** and possesses a **minimum value**.\n * If a < 0: Parabola opens **downward** and possesses a **maximum value**.\n * Worked Examples:\n * Example 1: Given f(x) = 2x^2 + 3x - 1,,a = 2 > 0 \implies parabola opens **upward**.\n * Example 2: Given f(x) = -3x^2 + 4x + 5,,a = -3 < 0 \implies parabola opens **downward**.\n * Guided Practice 5 - Opening Direction:\n 1. f(x) = 4x^2 + 2x - 3\n 2. f(x) = -2x^2 + 5x + 1\n 3. f(x) = -x^2 - 4x + 6\n 4. f(x) = 7x^2 - 3x - 2\n\n* **2. Vertex**\n * Definition: The turning point of the parabola (the absolute lowest point if opening upward, or the absolute highest point if opening downward).\n * Finding Vertex from Vertex Form f(x) = a(x - h)^2 + k:\n * Vertex is directly given by (h, k).\n * Worked Example: Given f(x) = 3(x - 2)^2 - 5 \implies \text{Vertex} = (2, -5).\n * Guided Practice 6:\n 1. f(x) = (x + 2)^2 - 1\n 2. f(x) = (x - 1)^2 - 4\n 3. f(x) = (x + 3)^2 - 4\n 4. f(x) = -(x - 2)^2 + 1\n * Finding Vertex from Standard Form f(x) = ax^2 + bx + c:\n * Step 1: Identify aandandb.\n * Step 2: Find x−coordinateusing-coordinate usingx = -\frac{b}{2a}.\n * Step 3: Substitute xbackintoback intof(x)tocalculatetheto calculate they-coordinate.\n * Worked Example: Given f(x) = x^2 - 8x + 15\n * Step 1: a = 1,,b = -8\n * Step 2: x = -\frac{-8}{2(1)} = -(-4) = 4\n * Step 3: f(4) = (4)^2 - 8(4) + 15 = 16 - 32 + 15 = -1\n * Result: Vertex is (4, -1).\n * Guided Practice 7:\n 1. f(x) = x^2 - 4x + 3\n 2. f(x) = x^2 + 6x + 5\n 3. f(x) = -x^2 + 4x + 1\n 4. f(x) = 2x^2 - 8x + 3\n\n* **3. Axis of Symmetry**\n * Definition: The vertical line dividing a parabola into two congruent, mirror-image halves; always passes directly through the vertex.\n * Formulas:\n * From Standard Form: x = -\frac{b}{2a}\n * From Vertex Form: x = h\n * Worked Examples:\n * Example 1: Given f(x) = x^2 - 6x + 5\n * Parameters: a = 1,,b = -6\n * Substitution: x = -\frac{-6}{2(1)} = 3 \impliesAxisofsymmetryisAxis of symmetry isx = 3\n * Example 2: Given f(x) = -(x + 1)^2 + 4\n * Rewrite x + 1asasx - (-1) \implies h = -1\n * Result: Axis of symmetry is x = -1\n * Example 3: Given f(x) = (x + 3)(x - 4)\n * Step 1 (Convert to standard form): f(x) = x^2 - 4x + 3x - 12 = x^2 - x - 12\n * Step 2 (Find x−coordinate):-coordinate):a = 1,,b = -1 \implies x = -\frac{-1}{2(1)} = \frac{1}{2}\n * Result: Axis of symmetry is x = \frac{1}{2}\n * Guided Practice 8 - Find Axis of Symmetry:\n 1. f(x) = x^2 - 8x + 5\n 2. f(x) = -x^2 - 4x + 3\n 3. f(x) = -(x - 3)^2 + 4\n 4. f(x) = -(x + 2)^2 + 9\n 5. f(x) = (x - 1)(x + 3)\n 6. f(x) = -(x - 1)(x - 5)\n\n* **4. y-Intercept**\n * Definition: Point where the parabola crosses the y−axis,evaluatedbysetting-axis, evaluated by settingx = 0.\n * Standard Form shortcut: For f(x) = ax^2 + bx + c,the, they−interceptis-intercept is(0, c).\n * Worked Examples:\n * Example 1: Given f(x) = 2x^2 - 5x + 7 \implies f(0) = 7 \implies y−interceptis-intercept is(0, 7).\n * Example 2: Given f(x) = -2(x + 2)^2 + 5\n * Step 1 (Substitute x = 0):):f(0) = -2(0 + 2)^2 + 5 = -2(2)^2 + 5 = -2(4) + 5 = -8 + 5 = -3\n * Step 2 (Write ordered pair): (0, -3)\n * Example 3: Given f(x) = 2(x + 1)(x - 3)\n * Step 1 (Substitute x = 0):):f(0) = 2(0 + 1)(0 - 3) = 2(1)(-3) = -6\n * Step 2 (Write ordered pair): (0, -6)\n * Guided Practice 9 - Find y-intercept:\n 1. f(x) = x^2 + 3x + 4\n 2. f(x) = 2x^2 - 5x - 6\n 3. f(x) = (x - 4)^2 - 9\n 4. f(x) = -(x + 4)^2 - 1\n 5. f(x) = -(x + 1)(x - 3)\n 6. f(x) = 2(x - 1)(x - 3)\n\n* **5. x-Intercepts**\n * Definition: Points where the parabola intersects or touches the x−axis,evaluatedbysetting-axis, evaluated by settingf(x) = 0$.

    • Worked Examples:

    • Example 1 (Standard Form factoring): Given f(x)=x2−5x+6f(x) = x^2 - 5x + 6

      • Step 1: Set x2−5x+6=0x^2 - 5x + 6 = 0

      • Step 2: Factor (x−2)(x−3)=0(x - 2)(x - 3) = 0

      • Step 3: x−2=0  ⟹  x=2x - 2 = 0 \implies x = 2; x−3=0  ⟹  x=3x - 3 = 0 \implies x = 3

      • Step 4: xx-intercepts are (2,0)(2, 0) and (3,0)(3, 0).

    • Example 2 (Factoring out GCF): Given f(x)=−2x2+10xf(x) = -2x^2 + 10x

      • Step 1: Set −2x2+10x=0-2x^2 + 10x = 0

      • Step 2: Factor out −2x  ⟹  −2x(x−5)=0-2x \implies -2x(x - 5) = 0

      • Step 3: −2x=0  ⟹  x=0-2x = 0 \implies x = 0; x−5=0  ⟹  x=5x - 5 = 0 \implies x = 5

      • Result: xx-intercepts are (0,0)(0, 0) and (5,0)(5, 0).

    • Example 3 (Vertex Form / Square Root Method): Given f(x)=(x+4)2−9f(x) = (x + 4)^2 - 9

      • Step 1: Set (x+4)2−9=0(x + 4)^2 - 9 = 0

      • Step 2: Isolate squared term   ⟹  (x+4)2=9\implies (x + 4)^2 = 9

      • Step 3: Take square root of both sides   ⟹  x+4=±3\implies x + 4 = \pm 3

      • Step 4: Solve for xx:

      • x+4=3  ⟹  x=3−4=−1x + 4 = 3 \implies x = 3 - 4 = -1

      • x+4=−3  ⟹  x=−3−4=−7x + 4 = -3 \implies x = -3 - 4 = -7

      • Result: xx-intercepts are (−1,0)(-1, 0) and (−7,0)(-7, 0).

    • Example 4 (Intercept Form / Zero Product Property): Given f(x)=(x+2)(x−4)f(x) = (x + 2)(x - 4)

      • Step 1: Set (x+2)(x−4)=0(x + 2)(x - 4) = 0

      • Step 2: Apply Zero Product Property: x+2=0x + 2 = 0 or x−4=0x - 4 = 0

      • Step 3: Solve for x  ⟹  x=−2x \implies x = -2 or x=4x = 4

      • Result: xx-intercepts are (−2,0)(-2, 0) and (4,0)(4, 0).

    • Guided Practice 10 - Find xx-intercepts:

    1. f(x)=x2−7x+12f(x) = x^2 - 7x + 12

    2. f(x)=x2−9x+20f(x) = x^2 - 9x + 20

    3. f(x)=(x−2)2−9f(x) = (x - 2)^2 - 9

    4. f(x)=(x+1)2−4f(x) = (x + 1)^2 - 4

    5. f(x)=(x+3)(x−1)f(x) = (x + 3)(x - 1)

    6. f(x)=(x−3)(x−5)f(x) = (x - 3)(x - 5)

  • 6. Minimum or Maximum Value

    • Definition: The extremum value of a quadratic function, which always equals the yy-coordinate of its vertex.

    • Conditions:

    • If a>0a > 0: Parabola opens upward →\rightarrow function has a minimum value.

    • If a<0a < 0: Parabola opens downward →\rightarrow function has a maximum value.

    • Worked Examples:

    • Example 1 (Standard Form): Given f(x)=2x2−12x+4f(x) = 2x^2 - 12x + 4

      • Step 1: a=2a = 2, b=−12b = -12, c=4c = 4

      • Step 2: Since a=2>0a = 2 > 0, parabola opens upward →\rightarrow minimum value exists.

      • Step 3: xx-coordinate of vertex: x=−−122(2)=124=3x = -\frac{-12}{2(2)} = \frac{12}{4} = 3

      • Step 4: yy-coordinate of vertex: f(3)=2(3)2−12(3)+4=18−36+4=−14f(3) = 2(3)^2 - 12(3) + 4 = 18 - 36 + 4 = -14

      • Result: Minimum value = −14-14

    • Example 2 (Vertex Form): Given f(x)=(x−2)2−5f(x) = (x - 2)^2 - 5

      • Step 1: Standard form expansion yields f(x)=x2−4x−1  ⟹  a=1>0f(x) = x^2 - 4x - 1 \implies a = 1 > 0.

      • Step 2: Vertex from vertex form is (2,−5)(2, -5).

      • Step 3: Since a>0a > 0, opening is upward →y\rightarrow y-coordinate is minimum value.

      • Result: Minimum value = −5-5

    • Example 3 (Vertex Form / Maximum): Given f(x)=−(x+1)2+4f(x) = -(x + 1)^2 + 4

      • Step 1: Standard form expansion yields f(x)=−x2−2x+3  ⟹  a=−1<0f(x) = -x^2 - 2x + 3 \implies a = -1 < 0.

      • Step 2: Vertex from vertex form is (−1,4)(-1, 4).

      • Step 3: Since a<0a < 0, opening is downward →y\rightarrow y-coordinate is maximum value.

      • Result: Maximum value = 44

    • Guided Practice 11 - Determine Minimum or Maximum Value:

    1. f(x)=(x−3)2−2f(x) = (x - 3)^2 - 2

    2. f(x)=−(x+2)2+5f(x) = -(x + 2)^2 + 5

    3. f(x)=2(x−1)2+4f(x) = 2(x - 1)^2 + 4

    4. f(x)=−3(x−4)2−1f(x) = -3(x - 4)^2 - 1

Domain and Range of Quadratic Functions

  • Domain

    • Definition: The set of all possible input xx-values.

    • Rule: For any quadratic function, xx can take any real number value.

    • Notation: Always (−∞,∞)(-\infty, \infty).

    • Example: For f(x)=x2−4x+3f(x) = x^2 - 4x + 3, Domain is (−∞,∞)(-\infty, \infty).

  • Range

    • Definition: The set of all possible output yy-values.

    • Rule: Depends strictly on the vertex yy-coordinate (kk) and the direction of opening:

    • If parabola opens upward (a>0a > 0): Range is y≥ky \ge k

    • If parabola opens downward (a<0a < 0): Range is y≤ky \le k

    • Worked Examples:

    • Example 1: Given f(x)=(x−2)2+3f(x) = (x - 2)^2 + 3

      • Step 1: Vertex is (2,3)(2, 3).

      • Step 2: Standard form f(x)=x2−4x+7  ⟹  a=1>0f(x) = x^2 - 4x + 7 \implies a = 1 > 0 (opens upward).

      • Step 3: Domain: (−∞,∞)(-\infty, \infty).

      • Step 4: Minimum yy-value is y=3  ⟹  y = 3 \implies Range: y≥3y \ge 3

    • Example 2: Given f(x)=−(x+1)2+5f(x) = -(x + 1)^2 + 5

      • Step 1: Vertex is (−1,5)(-1, 5).

      • Step 2: Standard form f(x)=−x2−2x+4  ⟹  a=−1<0f(x) = -x^2 - 2x + 4 \implies a = -1 < 0 (opens downward).

      • Step 3: Domain: (−∞,∞)(-\infty, \infty).

      • Step 4: Maximum yy-value is y=5  ⟹  y = 5 \implies Range: y≤5y \le 5

    • Guided Practice 12 - Determine Domain and Range:

    1. f(x)=(x−3)2+2f(x) = (x - 3)^2 + 2

    2. f(x)=−(x+2)2+6f(x) = -(x + 2)^2 + 6

    3. f(x)=2(x−1)2−4f(x) = 2(x - 1)^2 - 4

    4. f(x)=−(x−5)2−3f(x) = -(x - 5)^2 - 3

Graphing Quadratic Functions

  • Step-by-Step Procedure for Graphing:

    1. Identify the quadratic function.

    2. Determine the direction of opening.

    3. Find the vertex.

    4. Find the axis of symmetry.

    5. Find the yy-intercept.

    6. Find the xx-intercepts (if they exist).

    7. Plot the important points on the Cartesian plane.

    8. Use the axis of symmetry to identify corresponding symmetrical points.

    9. Draw a smooth, continuous U-shaped curve (parabola) through the points.

  • Comprehensive Graphing Worked Example:

    • Problem: Graph the quadratic function f(x)=x2−4x+3f(x) = x^2 - 4x + 3 on the Cartesian plane.

    • Step 1 (Identify coefficients): Compare with f(x)=ax2+bx+c  ⟹  a=1f(x) = ax^2 + bx + c \implies a = 1, b=−4b = -4, c = 3$.\n * Step 2 (Direction of opening): Since a = 1 > 0, the parabola opens **upward**.\n * Step 3 (Find vertex):\n * x−coordinate:-coordinate:x = -\frac{b}{2a} = -\frac{-4}{2(1)} = 2\n * y−coordinate:-coordinate:f(2) = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1\n * Vertex is (2, -1).\n * Step 4 (Axis of symmetry): Vertical line x = 2$.

    • Step 5 (Find yy-intercept): Set x=0  ⟹  f(0)=02−4(0)+3=3  ⟹  (0,3)x = 0 \implies f(0) = 0^2 - 4(0) + 3 = 3 \implies (0, 3).

    • Step 6 (Find xx-intercepts): Set f(x)=0  ⟹  x2−4x+3=0  ⟹  (x−1)(x−3)=0  ⟹  x=1f(x) = 0 \implies x^2 - 4x + 3 = 0 \implies (x - 1)(x - 3) = 0 \implies x = 1 or x=3  ⟹  (1,0)x = 3 \implies (1, 0) and (3,0)(3, 0).

    • Step 7, 8, & 9 (Plotting and Graphing):

      • Plot Vertex: (2,−1)(2, -1)

      • Plot xx-intercepts: (1,0)(1, 0) and (3,0)(3, 0)

      • Plot yy-intercept: (0,3)(0, 3)

      • Sketch smooth U-shaped parabola through all plotted points, ensuring symmetry about line x = 2$.\n\n![Graph of quadratic function f(x) = x^2 - 4x + 3 on the Cartesian plane with labelled points (0,3), (1,0), (2,-1), and (3,0)](https://assets.knowt.com/pdf-flow-prod/eba700a9-cfa1-4001-88b2-5e569f2c1d9e-figures/15.png)\n\n* Guided Practice 13 - Graphing Exercises:\n * Graph each function after finding opening direction, vertex, axis of symmetry, y−intercept,and-intercept, andx-intercepts:\n 1. f(x) = x^2 - 6x + 8\n 2. f(x) = (x - 2)^2 - 4\n 3. f(x) = (x + 2)(x - 2)$$