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Equations of Quadratic Functions from their Graphs (5.6)

Equations of Quadratic Functions from their Graphs (5.6). Working the other direction How many points determine a parabola?. First, a POD. Find the equation in slope-intercept form for the line passing through the points (1, 2) and (3, 5). First, a POD.

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Equations of Quadratic Functions from their Graphs (5.6)

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  1. Equations of Quadratic Functionsfrom their Graphs (5.6) Working the other direction How many points determine a parabola?

  2. First, a POD Find the equation in slope-intercept form for the line passing through the points (1, 2) and (3, 5).

  3. First, a POD Find the equation in slope-intercept form for the line passing through the points (1, 2) and (3, 5). Slope = (5-2)/(3-1) = 3/2 (y-2) = 3/2 (x-1) (y-5) = 3/2 (x-3) y-2 = 3/2 x - 3/2 y-5 = 3/2 x - 9/2 y = 3/2 x + 1/2 y = 3/2 x + 1/2

  4. We’ve found x and y values using given equations for parabolas Now, let’s find the equation using given points on the parabola. Try it with the points (-2, -11), (4, 13), and (6, 29). Start with y = ax2 + bx + c. Use the points to draft three equations in three variables: -11 = a(-2)2 + b(-2) + c 13 = a(4)2 + b(4) + c 29 = a(6)2 + b(6) + c

  5. Set them up as a system of equations Use your systems of equations skills to solve for a, b, and c. Here are the three equations to start with– take good notes now to follow the work. -11 = 4a -2b + c 13 = 16a + 4b + c 29 = 36a + 6b + c

  6. Plug values back in -11 = 4a -2b + c 13 = 16a + 4b + c 29 = 36a + 6b + c a = 1/2 b = 3 c = -7 So, the equation is y = (1/2)x2 + 3x - 7

  7. Regression You can do this on the calculators. • Hit Stat-Edit to find lists. Enter the x-values in L1 and the y-values in L2. • Hit Stat-Calc-QaudReg, then enter L1,L2. This will find the model that best fits those three points. The values for a, b, and c should be what we’ve already gotten. a = 1/2 b = 3 c = -7

  8. Matrices You can do this with matrices. • Enter the matrices. • Punch A-1(B) a = 1/2 b = 3 c = -7

  9. Try with a special case Use the points (0, 5), (2, 13), and (3, 26). Notice right off the bat that c = 5. Why? Use that to continue.

  10. Try with a special case Use the points (0, 5), (2, 13), and (3, 26). c = 5 4a + 2b + 5 = 13 9a + 3b + 5 = 26 a = 3 b = -2

  11. Try with another special case If one of the points given is a vertex, you need only two points, and you can use the vertex form of the equation. Try it. Find an equation of the parabola with vertex at (2, -5) and that goes through the point (3, 1). y - k = a(x - h)2 1 - (-5) = a(3 - 2)2 6 = a(1)2 a = 6 So, the equation is y + 5 = 6(x - 2)2.

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