Algebra 1 Study Guide: Quadratic Functions and Graphs

Graphing Quadratic Functions: Study Guide 8.1 - 8.5

Sketching Graphs of Quadratic Functions

  • General Instructions: Sketch the graph of each given quadratic function.
Example 1

  • f(x)=x2f(x) = x^2

  • A parabola with vertex at (0, 0), opening upwards.

Example 2

  • f(x)=−3x2f(x) = -3x^2

  • A parabola with vertex at (0, 0), opening downwards, stretched vertically by a factor of 3.

Example 3

  • f(x)=x2+6x+13f(x) = x^2 + 6x + 13

  • To find the vertex, complete the square or use x=−b/(2a)x = -b/(2a).

  • x=−6/(2∗1)=−3x = -6/(2*1) = -3

  • f(−3)=(−3)2+6(−3)+13=9−18+13=4f(-3) = (-3)^2 + 6(-3) + 13 = 9 - 18 + 13 = 4. Vertex at (-3, 4).

Example 4

  • f(x)=x2+8x+13f(x) = x^2 + 8x + 13

  • Find the vertex: x=−8/(2∗1)=−4x = -8/(2*1) = -4

  • f(−4)=(−4)2+8(−4)+13=16−32+13=−3f(-4) = (-4)^2 + 8(-4) + 13 = 16 - 32 + 13 = -3. Vertex at (-4, -3).

Example 5

  • f(x)=2x2−12x+14f(x) = 2x^2 - 12x + 14

  • Find the vertex: x=−(−12)/(2∗2)=12/4=3x = -(-12)/(2*2) = 12/4 = 3

  • f(3)=2(3)2−12(3)+14=18−36+14=−4f(3) = 2(3)^2 - 12(3) + 14 = 18 - 36 + 14 = -4. Vertex at (3, -4).

Example 6

  • f(x)=2x2−12x+21f(x) = 2x^2 - 12x + 21

  • Find the vertex: x=−(−12)/(2∗2)=12/4=3x = -(-12)/(2*2) = 12/4 = 3

  • f(3)=2(3)2−12(3)+21=18−36+21=3f(3) = 2(3)^2 - 12(3) + 21 = 18 - 36 + 21 = 3. Vertex at (3, 3).

Example 7

  • f(x)=(x+2)2−1f(x) = (x + 2)^2 - 1

  • This is in vertex form. Vertex at (-2, -1).

Example 8

  • f(x)=(x−4)2+2f(x) = (x - 4)^2 + 2

  • Vertex form. Vertex at (4, 2).

Example 9

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

  • Vertex at (-3, 1). Opens downward.

Example 10

  • f(x)=2(x−2)2−4f(x) = 2(x - 2)^2 - 4

  • Vertex at (2, -4). Opens upward and stretched by a factor of 2.

Example 11

  • y−2=(x+1)2y - 2 = (x + 1)^2

  • Rewrite as y=(x+1)2+2y = (x + 1)^2 + 2. Vertex at (-1, 2).

Example 12

  • y+2=(x−4)2y + 2 = (x - 4)^2

  • Rewrite as y=(x−4)2−2y = (x - 4)^2 - 2. Vertex at (4, -2).

Graphing Quadratic Functions: Vertex, Axis of Symmetry, Intercepts, Domain, and Range

Example 13
  • g(x)=3(x+1)(x+3)g(x) = 3(x + 1)(x + 3)
  • X-intercepts are (-1, 0) and (-3, 0).
  • Axis of symmetry is the average of the x-intercepts: x=(−1+−3)/2=−2x = (-1 + -3)/2 = -2
  • Vertex x-coordinate is -2. g(−2)=3(−2+1)(−2+3)=3(−1)(1)=−3g(-2) = 3(-2 + 1)(-2 + 3) = 3(-1)(1) = -3. Vertex at (-2, -3).
  • Domain: All real numbers.
  • Range: y≥−3y \geq -3
Example 14
  • y=−2(x+2)(x+4)y = -2(x + 2)(x + 4)
  • X-intercepts: (-2, 0) and (-4, 0).
  • Axis of symmetry: x=(−2+−4)/2=−3x = (-2 + -4)/2 = -3
  • Vertex x-coordinate is -3. y=−2(−3+2)(−3+4)=−2(−1)(1)=2y = -2(-3 + 2)(-3 + 4) = -2(-1)(1) = 2. Vertex at (-3, 2).
  • Domain: All real numbers.
  • Range: y≤2y \leq 2
Example 15

  • q(x)=x2+4x−5q(x) = x^2 + 4x - 5

  • Can be factored as (x+5)(x−1)(x + 5)(x - 1). X-intercepts at (-5, 0) and (1, 0).

  • Axis of symmetry: x=(−5+1)/2=−2x = (-5 + 1)/2 = -2

  • Vertex x-coordinate is -2. q(−2)=(−2)2+4(−2)−5=4−8−5=−9q(-2) = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -9. Vertex at (-2, -9).

  • Domain: All real numbers.

  • Range: y≥−9y \geq -9

Example 16
  • y=2x2−24x+64y = 2x^2 - 24x + 64
  • Axis of symmetry: x=−(−24)/(2∗2)=24/4=6x = -(-24)/(2*2) = 24/4 = 6
  • Vertex x-coordinate is 6. y=2(6)2−24(6)+64=72−144+64=−8y = 2(6)^2 - 24(6) + 64 = 72 - 144 + 64 = -8. Vertex at (6, -8).
  • Domain: All real numbers.
  • Range: y≥−8y \geq -8

Finding Zeros of Quadratic Functions

Example 17
  • y=x2−49xy = x^2 - 49x
    Factor y=x(x−49)y = x(x - 49). Zeros at x = 0 and x = 49.
Example 18
  • y=x3−x2−9x−9y = x^3 - x^2 - 9x - 9
    By factoring: y=x2(x−1)−9(x−1)=(x2−9)(x−1)=(x−3)(x+3)(x−1)y = x^2(x-1) - 9(x-1) = (x^2 - 9)(x-1) = (x-3)(x+3)(x-1). Zeros at -3, 1, and 3.

Using Zeros to Graph Quadratic Functions

Example 19
  • y=−x2−10x−16y = -x^2 - 10x - 16
  • Factor as y=−(x2+10x+16)=−(x+2)(x+8)y = -(x^2 + 10x + 16) = -(x + 2)(x + 8). Zeros at x = -2 and x = -8.
Example 20
  • h(x)=8x2−8h(x) = 8x^2 - 8
  • Factor as h(x)=8(x2−1)=8(x−1)(x+1)h(x) = 8(x^2 - 1) = 8(x - 1)(x + 1). Zeros at x = 1 and x = -1.

Writing Quadratic Functions in Standard Form

Example 21
  • Passes through (-4, 0), (3, 0), and (2, -18).
  • Since the zeros are -4 and 3, the function is of the form f(x)=a(x+4)(x−3)f(x) = a(x + 4)(x - 3).
  • Using the point (2, -18): −18=a(2+4)(2−3)=a(6)(−1)=−6a-18 = a(2 + 4)(2 - 3) = a(6)(-1) = -6a
  • Solving for a: a=−18/−6=3a = -18/-6 = 3
  • Thus, f(x)=3(x+4)(x−3)=3(x2+x−12)=3x2+3x−36f(x) = 3(x + 4)(x - 3) = 3(x^2 + x - 12) = 3x^2 + 3x - 36
Example 22
  • Vertex at (7, -3)
  • The equation is of the form f(x)=a(x−7)2−3f(x) = a(x - 7)^2 - 3
  • To find "a", you would need another point on the parabola (not provided in the transcript).