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explicitvogue

  • one year ago

Fan&medal to best help! What is the equation, in standard form, of a parabola that contains the following points? (-2,18), (0,2), (4,42) A. y=-2x^2-2x-3 B. y=-3x^2+2x-2 C. y=3x^2-2x+2 D. y=-2x^2+3x+2

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  1. anonymous
    • one year ago
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    By the points you've given the parabola would be opening down so your quadratic would definitely be negative.

  2. anonymous
    • one year ago
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    It actually opens up so its positive, oops I'll check my graphing calculator

  3. anonymous
    • one year ago
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    So definitely D

  4. ybarrap
    • one year ago
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    Using $$ (-2,18), (0,2), (4,42) $$ We can solve $$ y=ax^2+bx+c $$ for a, b and c. Using (-2,18), x=-2 and y=18: $$ 18=a-2^2+b(-2)+c=4a-2b+c $$ Using (0,2), x=0 and y=2: $$ 2=a0^2+b(0) + c=c $$ So now we know, c=2. Finally, for (4,42), x=4 and y=42: $$ 42=a4^2+b(4)+c=16a+4b+c $$ Putting this all together: $$ 18=4a-2b+2\\ 42=16a+4b+2 $$ Adding 2 times the 1st equation to the second we get: $$ 2(18)=2(4a-2b+2)\\ +\\ 42=16a+4b+2\\ =\\ 36+42=8a+16a-4b+4b+4+2\\ 78=24a+6\\ 24a=78-6=72\\ a=72/24=3 $$ So, $$ 18=4(3)-2b+2\\ 2b=12-18+2=-4\\ b=-2 $$ We have found that a=3,b=-2 and c=2: $$ y=3x^2-2x+2 $$ |dw:1444452119003:dw| Make sense?

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