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Quadratic Graphs

边玩边学

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课程笔记

Key Features of Quadratic Graphs

  • A quadratic graph has equation y = ax² + bx + c with a ≠ 0.
  • It is a smooth curve called a parabola, with a vertical line of symmetry.
  • If a > 0, the graph is u-shaped (minimum turning point).
  • If a < 0, the graph is n-shaped (maximum turning point).
  • The y-intercept is at (0, c).
  • The x-intercepts (roots) are solutions to ax² + bx + c = 0; there can be 0, 1, or 2 roots.
  • The turning point (vertex) is the minimum or maximum point.

Positive and negative quadratics

Positive and negative quadraticsO−8−6−4−22468−4−3−2−11234xyMax (0, 2)y = −x² +2 (a < 0)y = x² −2 (a > 0)Min (0, −2)

Sketching a Quadratic Graph

  • Draw axes and mark the y-intercept (0, c).
  • Find and mark the roots by solving ax² + bx + c = 0 (factorising, completing square, or quadratic formula).
  • Determine the shape: u-shaped if a > 0, n-shaped if a < 0.
  • Sketch a smooth curve through the intercepts, showing the turning point if known.
  • Label all intercepts and the turning point coordinates.

Key features of a quadratic graph

Key features of y = x² − 2x − 3O−4−2246−3−2−112345xy(−1, 0)(0, −3)(3, 0)y = x² −2x − 3Min (1, −4)

Finding the Turning Point by Completing the Square

  • Rewrite y = ax² + bx + c as y = a(x - p)² + q.
  • The turning point is at (p, q) (note sign change for p).
  • For y = (x - 3)² + 2, the minimum is (3, 2).
  • For y = (x + 3)² + 2, the minimum is (-3, 2).
  • The value of a does not affect the turning point coordinates but affects the shape.

Finding the Turning Point by Differentiation

  • Differentiate y = ax² + bx + c to get dy/dx = 2ax + b.
  • Set dy/dx = 0 and solve for x to find the x-coordinate of the turning point.
  • Substitute this x into the original equation to find the y-coordinate.
  • This method works for any quadratic and confirms whether it is a maximum or minimum.

Finding the Equation of a Quadratic from Its Graph

  • If the vertex (p, q) and one other point are known, use y = a(x - p)² + q.
  • Substitute the other point to find a.
  • If the roots (x₁, 0) and (x₂, 0) and one other point are known, use y = a(x - x₁)(x - x₂).
  • Substitute the other point to find a.
  • If a = 1, only the vertex or roots are needed.

Example: Sketching y = x² - 5x + 6

  • y-intercept: (0, 6) (c = 6).
  • Factorise: y = (x - 2)(x - 3) → roots at (2, 0) and (3, 0).
  • a = 1 > 0, so graph is u-shaped.
  • Sketch a smooth u-shaped curve through (0,6), (2,0), (3,0).

y = x² − 5x + 6

y = x² − 5x + 6O−2246810−1123456xy(0, 6)(2, 0)(3, 0)y = x² −5x + 6

Example: Sketching y = x² - 6x + 13

  • y-intercept: (0, 13).
  • Complete square: y = (x - 3)² + 4 → vertex at (3, 4) (minimum).
  • Since vertex is above x-axis and a > 0, there are no real roots.
  • Sketch u-shaped curve with vertex (3,4) and y-intercept (0,13).

y = x² − 6x + 13

y = x² − 6x + 13O51015−11234567xy(0, 13)y = x² −6x + 13Min (3, 4)

Example: Sketching y = -x² - 4x - 4

  • y-intercept: (0, -4).
  • Differentiate: dy/dx = -2x - 4; set to 0 → x = -2.
  • Substitute: y = -(-2)² - 4(-2) - 4 = 0 → vertex at (-2, 0) (maximum).
  • Only one root at x = -2 (touches x-axis).
  • Sketch n-shaped curve with vertex (-2,0) and y-intercept (0,-4).

y = −x² − 4x − 4

y = −x² − 4x − 4O−10−8−6−4−224−6−5−4−3−2−112xyMax (−2, 0)(−2, 0)y = −x² −4x − 4(0, −4)

Example: Finding Equation from Roots

  • Given roots at x = 2 and x = 3, and point (0, 24).
  • Use y = a(x - 2)(x - 3); substitute (0,24): 24 = a(-2)(-3) = 6a → a = 4.
  • Equation: y = 4(x - 2)(x - 3) or y = 4x² - 20x + 24.

y = 4(x − 2)(x − 3)

y = 4(x − 2)(x − 3)O510152025−112345xy(0, 24)(2, 0)(3, 0)y = 4(x −2)(x − 3)

Example: Finding Equation from Vertex

  • Given vertex at (9, -16) and point (2, 82).
  • Use y = a(x - 9)² - 16; substitute (2,82): 82 = a(2-9)² - 16 = 49a - 16 → 49a = 98 → a = 2.
  • Equation: y = 2(x - 9)² - 16 or y = 2x² - 36x + 146.

y = 2(x − 9)² − 16

y = 2(x − 9)² − 16O−202040608024681012xy(2, 82)y = 2(x −9)² − 16Min (9,−16)

幻灯片

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练习题

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  1. 1.What is the shape of the graph of y = x2 + 3x - 4?

    Easy
    • AU-shaped
    • BN-shaped
    • CStraight line
    • DS-shaped

  2. 2.The y-intercept of y = x2 - 5x + 6 is at (0, c). What is c?

    Easy
    • A6
    • B-5
    • C0
    • D-6

  3. 3.How many x-intercepts can a quadratic graph have?

    Easy
    • A0, 1 or 2
    • BAlways 2
    • CAlways 1
    • D0 or 2

  4. 4.The turning point of y = (x - 3)2 + 5 has coordinates:

    Easy
    • A(3, 5)
    • B(-3, 5)
    • C(3, -5)
    • D(-3, -5)

  5. 5.Complete the square: x2 + 6x + 11 = (x + a)2 + b. Find a and b.

    Medium
    • Aa = 3, b = 2
    • Ba = -3, b = 2
    • Ca = 3, b = 20
    • Da = -3, b = 20

  6. 6.The graph of y = -x2 + 4x - 3 has a maximum point. What is its x-coordinate?

    Medium
    • A2
    • B-2
    • C4
    • D-4

  7. 7.A quadratic graph has roots at x = 1 and x = 5. Which equation could represent it?

    Easy
    • Ay = (x - 1)(x - 5)
    • By = (x + 1)(x + 5)
    • Cy = (x - 1)(x + 5)
    • Dy = (x + 1)(x - 5)

  8. 8.The curve y = x2 - 6x + 13 has a turning point at (3, 4). How many x-intercepts does it have?

    Medium
    • A0
    • B1
    • C2
    • DCannot be determined

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