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anonymous
 one year ago
using these three coordinates write an equation for the parabola in these forms
anonymous
 one year ago
using these three coordinates write an equation for the parabola in these forms

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437628694097:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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) \[0=a(0r_1)(0r_2 ) / x=0 , y=0 \] \[0=ar_1 r_2.............(I)\] 2. Point (20,12) \[12=a(20r_1)(20r_2)=400a20ar_2 20ar_1+ar_1r_2\] But you know from (I) that ar1r2=0 \[12=400a20ar_2 20ar_1.........(II)\] 3. Point (40,0) \[0=a(40r_1)(40r_2)=1600a40ar_2 40ar_1+ar_1r_2\] And again we know from (I) that ar1r2=0 \[0=1600a40ar_2 40ar_1\] dividing by 2 \[0=800a20ar_2 20ar_1........(III)\] Now solving the equation system (II) and (III) \[12=400a20ar_2 20ar_1.........(II)\] \[0=800a20ar_2 20ar_1........(III)\] 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 \[y=0.03(x)(x40)\] Now do the same for your second expression

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437645525075:dw parabola opens down a is negative vertex (20,12) line of symmetry x = 20 focus at (20, 6)
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