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anonymous

  • one year ago

using these three coordinates write an equation for the parabola in these forms

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  1. anonymous
    • one year ago
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    |dw:1437628694097:dw|

  2. anonymous
    • one year ago
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    You know the parabola goes through those 3 points, so those points are solutions for the parabola, being a solution means that replacing the point will make the equation right (or true). For your first expression 1. Point (0,0) \[0=a(0-r_1)(0-r_2 ) / x=0 , y=0 \] \[0=ar_1 r_2.............(I)\] 2. Point (20,12) \[12=a(20-r_1)(20-r_2)=400a-20ar_2 -20ar_1+ar_1r_2\] But you know from (I) that ar1r2=0 \[12=400a-20ar_2 -20ar_1.........(II)\] 3. Point (40,0) \[0=a(40-r_1)(40-r_2)=1600a-40ar_2 -40ar_1+ar_1r_2\] And again we know from (I) that ar1r2=0 \[0=1600a-40ar_2 -40ar_1\] dividing by 2 \[0=800a-20ar_2 -20ar_1........(III)\] Now solving the equation system (II) and (III) \[12=400a-20ar_2 -20ar_1.........(II)\] \[0=800a-20ar_2 -20ar_1........(III)\] we get a=-0.03, we also know for (I) that one of r1 & r2 should be 0, we pick one randomly and replacing a and r1=0 in (II) or (III) gives us the value of r2=40 So your first expression should be \[y=-0.03(x)(x-40)\] Now do the same for your second expression

  3. triciaal
    • one year ago
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    |dw:1437645525075:dw| parabola opens down a is negative vertex (20,12) line of symmetry x = 20 focus at (20, 6)

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