1 / 24

Precalc – 2.1 - Quadratics

Precalc – 2.1 - Quadratics. What do we already know?. Representations. ALGEBRA. POLYNOMIAL FORM. STANDARD FORM. FACTORED FORM. y = ax 2 + bx + c. y = a(x - h) 2 + k. y = ( x + a)(x + b ) . ORDERED PAIRS. GRAPH. Parabola Graph Vocabulary.

gefen
Download Presentation

Precalc – 2.1 - Quadratics

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. Precalc – 2.1 - Quadratics

  2. What do we already know?

  3. Representations ALGEBRA POLYNOMIAL FORM STANDARD FORM FACTORED FORM y = ax2 + bx + c y = a(x - h)2 + k y = (x + a)(x + b) ORDERED PAIRS GRAPH

  4. Parabola Graph Vocabulary • Axis of symmetry: the line that splits the parabola in half • Vertex: the points where the axis of symmetry intersects the parabola; also, either the highest or the lowest point on the graph • Opens upward: graph will look like a U • Opens downward: graph will look like an upside down U

  5. Algebraic  Graphic representation • Moving between algebraic and graphical representations is difficult because different representations give you different types of information • Polynomial form: vertex, opens up or down, zeros • but through the use of formulas • Standard form: vertex, opens up or down, skinny/wide • Factored form: zeros • note: not every parabola has zeros

  6. Standard Form  Graph • You already know how to do this! • Graph: y = -2(x – 2)2 + 4 • Vertex of the equation y = a(x – h)2 + k

  7. Practice • What are the vertices of these equations? • y = 3(x – 5)2 + 9 • y = 4(x + 9)2 + 7 • y = (x – 7)2 • y = x2 + 7 • y = x2 • y = 2(x + 10)2

  8. Standard  Graph • How will we know where its zeros are? • We can only get that information from the polynomial or factored form • *As always, we set y=0, solve for x

  9. Moving between algebraic forms POLYNOMIAL FORM STANDARD FORM FACTORED FORM y = ax2 + bx + c y = a(x - h)2 + k y = (x + a)(x + b) Notice that to move from standard to factored forms, we have to pass through the polynomial form

  10. Standard  Polynomial • Expand the squared section and combine like terms • Ex: y = 2(x – 1)2 – 8 y = 2(x – 1)2 – 8 y = 2(x2 – 2x + 1) – 8 y = 2x2 – 4x + 2 – 8 y = 2x2 – 4x – 6

  11. Practice • Turn these into polynomial form • y = 3(x – 5)2 + 9 • y = 4(x + 9)2 + 7 • y = (x – 7)2 • y = 2(x + 10)2

  12. Polynomial  Factored • Factor! • y = 2x2 – 4x – 6 hm… remember how to factor?

  13. How to factor: • Given: f(x)=ax2+bx+c • Find two numbers that: • Add up to b • Multiply out to axc • Rewrite the equation with those numbers • Pair up your terms and find common factors • Factor the pairs • Find the common factors of the factored pairs • Rewrite as a product of two binomials

  14. Practice • Factor these equations • y = 2x2– 4x – 6 • y = 2x2+3x-5 • y = 6x2+5x+1

  15. Reminder: • Why did we want to change into polynomial and factored forms? • Oh yes – because we can’t get the zeros from standard form • We need the zeros to graph a parabola • Okay! So let’s get those zeros.

  16. Factored  Graph • y = (x + a)(x + b) • The zeros are: • In our example: y = (2x + 2)(x – 3) 2x + 2 = 0 x – 3 = 0

  17. Practice • Graph these functions (vertex from vertex form, zeros from factored form) • y = 2(x – 1)2 – 8 • y = -(x + 4)2 + 9 • y = 4(x – 2)2 - 4

  18. Polynomial  Graph • We didn’t have to factor the equation to get the zeros; there is one other option • 0 = 2x2 – 4x – 6 • Quadratic Formula! Tells the values of x that make the function zero.

  19. Polynomial  Graph • We can also find the vertex • Given the equation y = ax2 + bx + c, the vertex will be: • This comes from calculus

  20. Practice • Graph these functions (zeros from quadratic formula, zeros from –b/2a) • y = 2x2 – 4x – 6 • y = -x2 – 8x – 7 • y = 4x2 – 16x + 12

  21. ALGEBRA POLYNOMIAL FORM STANDARD FORM FACTORED FORM y = ax2 + bx + c y = a(x - h)2 + k y = (x + a)(x + b) GRAPH Put it together: how do we move between all these different representations of quadratic functions?

  22. Exceptions • Why will we have difficulty with a function that has a graph like this?

  23. Graph  Vertex Form

  24. HOMEWORK! • Page 208 #1-8, 13, 18, 20, 23, 24, 26

More Related