1 / 9

MTH 065 Elementary Algebra II

MTH 065 Elementary Algebra II. Chapter 11 Quadratic Functions and Equations Section 11.7 More About Graphing Quadratic Functions. Forms of Quadratic Functions. Standard form … f(x) = ax 2 + bx + c Graphing form … f(x) = a(x – h) 2 + k. How to convert between these forms ….

kendall-huff
Télécharger la présentation

MTH 065 Elementary Algebra II

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. MTH 065Elementary Algebra II Chapter 11 Quadratic Functions and Equations Section 11.7 More About Graphing Quadratic Functions

  2. Forms of Quadratic Functions • Standard form … f(x) = ax2 + bx + c • Graphing form … f(x) = a(x – h)2 + k How to convert between these forms …

  3. Graphing Form  Standard From • Square the binomial, remove grouping symbols, and combine like terms … f(x) = 2(x – 4)2 + 5 f(x) = 2(x2 – 8x + 16) + 5 f(x) = 2x2 – 16x + 32 + 5 f(x) = 2x2 – 16x + 37

  4. Standard Form  Graphing From • Factor a out of the terms w/ the variable, complete the square, factor, and add the constants … f(x) = 2x2 – 16x + 37 f(x) = 2(x2 – 8x) + 37 f(x) = 2(x2 – 8x + 42) + 37 – 32 f(x) = 2(x – 4)2 + 5

  5. Locating the Vertex from the Standard From f(x) = 2x2 – 16x + 37 f(x) = 2(x2 – 8x) + 37 f(x) = 2(x2 – 8x + 42) + 37 – 32 f(x) = 2(x – 4)2 + 5 • To get to the 4 (the x-coordinate of the vertex) … • Start with the opposite of the linear coefficient (b) … • Divide by the quadratic coefficient (a) … • Divide by 2 • i.e.

  6. Locating the Vertex from the Standard From f(x) = 2x2 – 16x + 37 f(x) = 2(x2 – 8x) + 37 f(x) = 2(x2 – 8x + 42) + 37 – 32 f(x) = 2(x – 4)2 + 5 • To get to the 5 (the y-coordinate of the vertex) … • Start with the constant term (c) … • Subtract a times the square of the x-coordinate … • i.e.

  7. Locating the Vertex f(x) = ax2 + bx + c Vertex:

  8. The y-intercept • Let x = 0 f(x) = ax2 + bx + c f(0) = a(0)2 + b0 + c f(0) = c • Therefore, the constant term is the y-intercept.

  9. The x-intercepts • Let f(x) = 0 f(x) = ax2 + bx + c 0 = ax2 + bx + c … solve for x (if possible) … • Therefore, there may be 0, 1, or 2 x-intercepts.

More Related