logging in or signing up Quadratic functions Melvin6375 Download Post to : URL : Related Presentations : Let's Connect Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Copy embed code: Embed: Flash iPad Dynamic Copy Does not support media & animations Automatically changes to Flash or non-Flash embed WordPress Embed Customize Embed URL: Copy Thumbnail: Copy The presentation is successfully added In Your Favorites. Views: 333 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: February 13, 2012 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript PowerPoint Presentation: Quadratic Functions and ModelsPowerPoint Presentation: If a > 0 the parabola opens up and the larger the a value the “narrower” the graph and the smaller the a value the “wider” the graph. If a < 0 the parabola opens down and the larger the a the “narrower” the graph and the smaller the a the “wider” the graph.PowerPoint Presentation: vertical shift, moves graph vertically by k horizontal shift, moves graph horizontally by h Determines whether the parabola opens up or down and how “wide” it is We need to algebraically manipulate this to look like the form above. We’ll do this by completing the square. Add a number here to make a perfect square Subtract it here to keep things equal (can’t add a number without compensating for it and we don’t want to add it to the other side because of function notation) 9 9 This will factor into ( x +3)( x +3) so we can express it as something squared and combine the -1 and -9 on the end. The graph of this function is a parabola Let’s look at a quadratic function and see if we can graph it.PowerPoint Presentation: right 3 down 10 We started with and completed the square to get it in the format to be able to graph using transformations. We can take a general quadratic equation and do this to find a formula for the vertex. This is done in your book at the bottom of page 295 by the number 2. What we find from doing this is on the next slide.PowerPoint Presentation: The x value of the vertex of the parabola can be found by computing The y value of the vertex of the parabola can be found by substituting the x value of the vertex in the function and finding the function value. Let’s try this on the one we did before: 1 (1) (-6) The vertex is then at (3, -10)PowerPoint Presentation: (3, -10) Let’s plot the vertex: Since the a value is positive, we know the parabola opens up. The parabola will be symmetric about a vertical line through the vertex called the axis of symmetry. Let’s find the y intercept by plugging 0 in for x . So y intercept is (0, -1) The graph is symmetric with respect to the line x = 3 so we can find a reflective point on the other side of the axis of symmetry. (0, -1) (6, -1) We can now see enough to graph the parabolaPowerPoint Presentation: (3, -10) Let’s look at another way to graph the parabola starting with the vertex: We could find the x intercepts of the graph by putting f ( x ) (which is the y value) = 0 This won’t factor so we’ll have to use the quadratic formula. So x intercepts are (6.2, 0) and (- 0.2, 0)PowerPoint Presentation: A mathematical model may lead to a quadratic function. Often, we are interested in where the function is at its minimum or its maximum. If the function is quadratic the graph will be a parabola so the minimum (if it opens up) will be at the vertex or the maximum (if it opens down) will be at the vertex. We can find the x value of the vertex by computing We could then sub this value into the function to find its minimum or maximum value.PowerPoint Presentation: DEMAND EQUATION The price p and the quantity x sold of a certain product obey the demand equation: This is the real world domain. The equation doesn’t make sense if the quantity sold is negative ( x < 0) and it doesn't make sense if the price is negative (if x > 400) Express the revenue R as a function of x . Revenue is the amount you bring in, so it would be how much you charge (the price p ) times how many you sold (the quantity x )PowerPoint Presentation: This is a quadratic equation and since a is negative, its graph is a parabola that opens down. It will have a maximum value then at the y value of the vertex. What is the revenue if 100 units are sold? What quantity x maximizes revenue? Since the revenue function is maximum at the vertex, we'll want to find the x value of the vertex to answer this. What is the maximum revenue? This would be the y value of the vertexPowerPoint Presentation: DEMAND EQUATION The price p and the quantity x sold of a certain product obey the demand equation: What price should the company charge to receive maximum revenue? Since we just found that the quantity to achieve maximum revenue was 200, we can substitute this in the price equation to answer this question. You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.