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anonymous

  • 5 years ago

I need help with parabolas and solving quadratic-linear systems. 8th grade algebra one.

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  1. amistre64
    • 5 years ago
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    whats the q?

  2. amistre64
    • 5 years ago
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    from the sound of the statement you made, it seems like you need help finding where quads and lines intersect each other...

  3. anonymous
    • 5 years ago
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    Yes but I don't understand the step in the process that is xorigin = -)b/2a) = 0. Does that have any significance and does it always equal 0?

  4. amistre64
    • 5 years ago
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    do you know the generic form for a quadratic equation? ax^2 +bx + c ?

  5. amistre64
    • 5 years ago
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    And do you know the generic shape for a quadratic equation that is graphed? it looks like a "U" right?

  6. anonymous
    • 5 years ago
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    I didn't know the generic form for a quadratic equation but I did know about the parabola. But these types of problems end up to be a parabola with a line intersecting it.

  7. amistre64
    • 5 years ago
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    the formula (-b/2a) is gotten from the generic form of the quadratic equation; where a = some number; b = some number; and c = some number... for example: 2x^2 +3x +7 is a quadratic equation with a=2; b=3; and c = 7 the form -b/2a is the formula to find the "axis of symmetry" of the parabole; and it is the value of "x" that that "U" shape strats to bend around itself and go back up again; in this equation tha twould be: -3/2(2) ; which equals -3/4

  8. amistre64
    • 5 years ago
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    if the parabold is positioned at the origin; then the form (-b/2a) will always be zero; but the parabola is not always positioned at the otigin :)

  9. amistre64
    • 5 years ago
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    you gotta read thru the typos :)

  10. anonymous
    • 5 years ago
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    Oh I get it! Thank you so much! Alright, so let's say I have two equations; y = x^2 and y = x +2 What would a and b and c be equal to?

  11. anonymous
    • 5 years ago
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    thanks amistre for explaining it to her. sorry about the paula got sidetracked.

  12. anonymous
    • 5 years ago
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    @abigail - it's fine

  13. amistre64
    • 5 years ago
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    for the quadratic part; a = 1; b = 0 and c = 0; the linear part has different "constants" to it tho. what I think you are looking to find is the points that x^2 and x+2 have in common like this:

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  14. amistre64
    • 5 years ago
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    abigail: :) is alright, you can still help if youd like ;)

  15. anonymous
    • 5 years ago
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    Ya that's what I need to find ;) so when I'm solving this should I write the formula would -b/2a be equal to 0?

  16. amistre64
    • 5 years ago
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    when we wnat to know the points that grpah have in common; we make their equations equal like this: x^2 = x+2 ; next we put everything to one side x^2 -x -2 = 0 ; next we solve for the values of "x" that make this equation equal 0, and we do that easily by the quadratic formula. Or by factoring the equation. the quadratic formula is: -b/2a +- sqrt(b^2 -4(a)(c))/2a

  17. amistre64
    • 5 years ago
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    x^2 -x -2; a=1, b=-1, c=-2 --1/2(1) +- sqrt((-1)^2 - 4(1)(-2))/2(1) 1/2 +- sqrt(1+8)/2 1/2 +- sqrt(9)/2 1/2 +- 3/2 x = 1/2 + 3/2 = 4/2 = 2 AND x = 1/2 - 3/2 = -2/2 = -1 the points they have in common are going to be: (-1,y1) (2,y2)

  18. amistre64
    • 5 years ago
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    we find y1 and y2 by using the values for x that we found into one of the equations: x^2 is a good choice if you know your squares :) (-1)^2 = 1 (-1,1) is a point they share in common (2)^2 = 4 (2,4) is the other point they share in common

  19. anonymous
    • 5 years ago
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    Gotchya :D Thanks sooo much!

  20. anonymous
    • 5 years ago
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    id say you covered everything amistre i have nothing to input on this lol

  21. amistre64
    • 5 years ago
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    :) mighta covered it all, but explaining it good enough is another matter :)

  22. anonymous
    • 5 years ago
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    thats soo the truth either way well done!

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