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Quadratic Function By Mr. Keung/Ms. Keung

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Abstract Currently we are working on quadratic functions so we have designed a PowerPoint lesson which has been used in our classrooms with success. We start with the topic ‘Quadratic Function’ and later extend to the quadratic equations. Our students are 8th graders in the John D O’Bryant and 9th graders in the Snowden High, and we both use the same textbook, Algebra 1 (Glencoe). Essentially we just hook up the Infocus with our laptop and conduct the lesson. We both have a couple of old computers in our rooms, therefore some students can go online together with us while we are doing the online exploration part. In this particular lesson we are using the gizmo, ‘Roots of Quadratic Function’, from the Explorelearning.com.

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Quadratic Function (y = ax 2 + bx + c) a, b, and c are called the coefficients. The graph will form a parabola. Each graph will have either a maximum or minimum point. There is a line of symmetry which will divide the graph into two halves.

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y = x 2 a = 1, b = 0, c = 0 Minimum point (0,0) Axis of symmetry x=0 y=x 2

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What happen if we change the value of a and c ? y=3x 2 y=-3x 2 y=4x 2 +3 y=-4x 2 -2

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Conclusion (y = ax 2 +bx+c) When a is positive, When a is negative, When c is positive When c is negative the graph concaves downward. the graph concaves upward. the graph moves up. the graph moves down.

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What happens if b varies? Explore http://www.explorelearning.com/index. cfm?method=cResource.dspView&Reso urceID=154 http://www.explorelearning.com/index. cfm?method=cResource.dspView&Reso urceID=154 Describe the changes in your own words.

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Solving Quadratic Functions (ax 2 + bx + c = 0) Since y = ax 2 + bx +c, by setting y=0 we set up a quadratic equation. To find the solutions means we need to find the x-intercept. Since the graph is a parabola, there will be two solutions.

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To solve quadratic equations (graphing method) X 2 - 2x = 0 To solve the equation, put y = x 2 -x into your calculator. Find the x intercept. Two solutions, x=0 and x=2. y=x 2 -2x

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Find the Solutions y=x 2 -4 y=x 2 +2x-15 y=-x 2 +5y=-x 2 -1

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Find the solutions y=x 2 +2x+1 y=-x 2 +4x-1

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Observations Sometimes there are two solutions. Sometimes there is only one solution. Sometimes it is hard to locate the solutions. Sometimes there is no solution at all.

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Other Methods By factoring By using the quadratic formula

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The End

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