1 / 20

Quadratic Equations

Quadratic Equations. Unit 2 English Casbarro. Section 1- Graphing Quadratic Equations Quadratic Equations are equations in the form The graphs of quadratic equations are called parabolas. If a is positive, the parabola points up, ; if a is negative, the parabola points down,

aglasco
Télécharger la présentation

Quadratic Equations

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. Quadratic Equations Unit 2 English Casbarro

  2. Section 1-Graphing Quadratic EquationsQuadratic Equations are equations in the form The graphs of quadratic equations are called parabolas. If a is positive, the parabola points up, ; if a is negative, the parabola points down, The following graphs show the vertex (–3, 0) of one parabola (which is the only solution), and the solutions (1, 0) and (4, 0) of another parabola. The third parabola has imaginary solutions.

  3. Graphing from Standard Form Standard and Vertex form follow the same rules. There are 3 things you always find: • The vertex • The y-intercept • The matching point

  4. Standard Form: • 1. Finding the vertex: • You must use the formula: ,so that you can find the axis of symmetry and the x-value of the vertex. • Next, you substitute the x into the equation to find the y-value of the vertex. • So the vertex is (1,5)

  5. Vertex Form: • From the previous slide, we found the vertex of the parabola, using the formula: • and substituting the x value into the equation to find • the y-value of the vertex. You found that the vertex is (1,5). • The “a” from your equation is the same “a” in the “a” in standard form. So, • The equation in vertex form is:

  6. Vertex form without completing the square Now that you can find the vertex of any quadratic function, it is easy to put the function in vertex form. Ex. Put in vertex form. Find So, Now, So, the vertex is (1,8) and a = 3. The equation in vertex form is:

  7. Section 7: Quadratic Inequalities You have graphed in all 3 forms of Quadratic Equations: • Standard Form • Vertex Form • Solutions (factored) Form Now you will graph the parabolas, then shade for the inequalities.

  8. Graphing You must still find the 3 points to graph: • The vertex • The y-intercept • The matching point

  9. Example 1: First, we find the vertex:

  10. Next, we find the y-intercept: • Substitute 0 into the inequality: • The y-intercept is (0,3).

  11. The matching point • The vertex is: • The y-intercept is: • The matching point: • Graph these 3 points

  12. Shading • The inequality was: We always shade “above” the graph for The equal sign below it means that the line is solid.

  13. Graph

  14. Using the calculator • Go to Y= • Use the direction buttons to move as far to the left as you can • Use the “Enter” button to change the type of display • is represented by the symbol • Type in the quadratic inequality • Graph the inequality: the shading will be automatic.

  15. Section 8: Modeling • The vertex form of a quadratic equation is: y + a(x – h)2 + k • You can use this form to find equations if you are given points

  16. Given the vertex and a point Ex. 1: (#10 in your book) Vertex: (2, –1) Point: (4, 3) The equation is y = a(x – h)2 + k This means that you are given, as the question, 4 out of the 5 variables in the general equation. (h = 2, k = –1) (x = 4, y = 3)

  17. Fill in the general equation y = a(x – h)2 + k 3 = a(4 – 2)2 + (–1) (substitute h, k, x, and y) 3 = a(2)2 – 1 3 = 4a – 1 (add 1 to both sides) 4 = 4a (divide both sides by 4) a = 1

  18. Fill in the equation with h, k, a The general equation is : y = a(x – h)2 + k (h, k)  (2, –1) a = 1 So the specific equation is: y = 1(x – 2)2 – 1

  19. You try this: Vertex: (4, 5) Point: (8, –3)

  20. Solution: y = a(x – h)2 + k Substitute h, k, x, and y –3 = a(8 – 4)2 + 5 –3 = a(4)2 +5 Subtract 5 from both sides –8 = 16a Divide both sides by 16 a = -1/2 So the specific equation is y = –1/2(x – 4)2 + 5

More Related