1 / 8

Understanding Vertex Form and Transforming Quadratic Functions

This lesson focuses on the vertex form of quadratic functions, represented as y = a(x - h)² + k. Students will learn how to identify the vertex (h, k), determine if the parabola opens up or down based on the value of 'a', and find the y-intercept. Additionally, the lesson covers converting from vertex form to polynomial form through the process of expanding and combining like terms. Sample problems will guide students in these transformations, along with homework assignments to reinforce the concepts learned.

tammy
Télécharger la présentation

Understanding Vertex Form and Transforming Quadratic Functions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Unit #3: Quadratics5-3: Translating Parabolas Essential Question: How is the vertex form used in transforming a quadratic function?

  2. 5-3: Translating Parabolas • Yesterday, we explored quadratic equations in polynomial form: y = ax2 + bx + c • Two other forms which are used are x-intercept (factored) form and vertex (transformation) form. • Vertex form of a quadratic function is an equation in the form: y = a(x – h)2 + k • Just like in polynomial form: • If a is positive, the graph opens up • If a is negative, the graph opens down

  3. 5-3: Translating Parabolas • y = a(x – h)2 + k • The vertex is found at (h, k) • h: Set what’s inside the parenthesis = 0 and solve for x • k: Take the number that’s being added to the end of the function. • Example • Find the vertex of y = (x + 3)2 – 1 • Because a = 1 (nothing in front of the parenthesis): • The graph opens up. • x + 3 = 0 → subtract 3 on each side →x = -3 • h = -3, k = -1 • The vertex is at (-3, -1).

  4. 5-3: Translating Parabolas • Determine the vertex of the quadratic function and whether the graph opens up or down. • y = -3(x + 2)2 + 4 • Opens: • Vertex: • y = 2(x – 2.5)2 – 5.5 • Opens: • Vertex: • y = -½(x – 2)2 • Opens: • Vertex: Down (-2, 4) Up (2.5, -5.5) Down (2, 0)

  5. 5-3: Translating Parabolas • To find the y-intercept: • Simply substitute “0” in for x and simplify • Example: y = -2(x – 3)2 + 4 • y = -2(0 – 3)2 + 4 • = -2(-3) 2 + 4 • = -2(9) + 4 • = -18 + 4 • = -14 • The y-intercept is at (0, -14)

  6. 5-3: Translating Parabolas • To convert into polynomial form: • FOIL the parenthesis • DISTRIBUTE (if necessary) the number outside • COMBINE like terms • Example: y = -2(x – 3)2 + 4 • y = -2(x – 3)2 + 4 • = -2(x – 3)(x – 3) + 4 • = -2(x2 – 6x + 9) + 4 • = -2x2 + 12x – 18 + 4 • y = -2x2 + 12x – 14

  7. 5-3: Translating Parabolas • Determine the y-intercept of the quadratic function and convert the function into polynomial form. • y = -3(x + 2)2 + 4 • y-intercept: • Polynomial form: • y = (x – 3)2 – 4 • y-intercept: • Polynomial form: • y = -½(x – 2)2 • y-intercept: • Polynomial form: (0, -8) y = -3x2 – 12x – 8 (0, 5) y = x2 – 6x + 5 (0, -2) y = -½x2 + 2x – 2

  8. 5-3: Translating Parabolas • Assignment • Page 251 • Problems 1-11, odd • Ignore the directions!!! • Tell me: • Whether the graph opens up or opens down • The vertex • The y-intercept • The function written in polynomial form Note: You may want to do part (d) before part (c) • Tomorrow, Practice with graphing • Friday, Quiz on 5-1 through 5-3

More Related