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clamin

  • one year ago

PLEASE HELP!!! MEDAL!!! Please show me the x and y value of y=-x^2+2x-2

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  1. sepeario
    • one year ago
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    Okay let's solve for x first. Have you heard of the quadratic formula?

  2. clamin
    • one year ago
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    Yes. y=ax^2+bx+c @sepeario

  3. anonymous
    • one year ago
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    you can break it up like this :) well, you're looking for to this formula : \[x=\frac{ b \pm \sqrt{b ^{2} -4ac}} {2a }\] now let's solve it :) -x^2+2x-2 : this is your equation, now let's break down a little bit: -x^2 = a=(-1): when it's x, you can put it as one to easy to calculate, plus, it make sense, you can use any other real number but the results it not so beautiful... okay, next: 2x = b -2 = c Now we know all the numbers and letters: \[y= \frac{ 2\pm \sqrt{2^{2}-4(-1)(-2)} }{ 2(-1) }\] \[y=\frac{ 2\pm \sqrt{4-8} }{ -2 }\] Since \[\sqrt{4-8} = \sqrt{-4}\] which doesn't make sense, so it will be false! which now we know that it will be NO SOLUTION! :) hope it help :)

  4. clamin
    • one year ago
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    This helps a little but I'm looking for the x and y value which I struggle a lot to find..@leong

  5. clamin
    • one year ago
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    @Leong

  6. anonymous
    • one year ago
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    huh?

  7. anonymous
    • one year ago
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    you can't look for x and y when you know that's it false, you can try your formula which is (a+b)^2= (a^2+2ab+b^2), I did that and it give me very ugly numbers.....

  8. anonymous
    • one year ago
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    not every equations have numbers or results, since the equations is kinda false itself. Like the one you give me, I can show you if you wanted by using others quadratic formula, but trust me, you will ending up with ugly numbers or false :)

  9. clamin
    • one year ago
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    Its for the graph

  10. freckles
    • one year ago
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    what do you mean to show you the x and y value?

  11. freckles
    • one year ago
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    x and y are variables they vary they don't just take on one value

  12. sepeario
    • one year ago
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    i think he/she means solve for x and y.

  13. Nnesha
    • one year ago
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    x and y intercepts ? @clamin ?

  14. Nnesha
    • one year ago
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    if yes then use \(\huge\color{reD}{\rm b^2-4ac}\) `Discriminant` if ` b^2-4ac > 0` then there are 2 real zeros if ` b^2-4ac = 0` then there is one real root if ` b^2-4ac < 0` then you will get two complex roots (no -x-intercept) to find y-intercept substitute x for 0 solve for y

  15. phi
    • one year ago
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    If you want some (x,y) pairs so that you can graph the equation \[ y=−x^2+2x−2 \] I would start by picking some small x values (they are easier to figure out) for example, when x is 0 we find y by doing this: \[ y=−(0)^2+2⋅0−2 \] hopefully you know that simplifies to y=-2 so (0,-2) is one point

  16. phi
    • one year ago
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    let's try x=-1: \( y=−(−1)^2+2⋅−1−2\) notice that when we replaced x with -1 in -x^2 we put in parens -(-1)^2 - (-1)^2 means -1* (-1)^2 or -1* (-1*-1) or -1*1 and finally -1 we get y=−1−2−2=−5 so (-1,-5) is another point

  17. phi
    • one year ago
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    now try x=1 y= -(1)^2 +2*1 -2 y= -(1*1) + 2 -2 y= -1 so (1,-1) is a point try x=2 y= -(2*2) + 2*2 -2 = -4+4-2 = -2 (2,-2) x=3 y= -(3*3) +2*3 -2 = -9 +6-2 = -5 (3,-5)

  18. clamin
    • one year ago
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    @phi how come the 4 is negative?

  19. clamin
    • one year ago
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    @nnesha @freckles @sepeario @leong

  20. phi
    • one year ago
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    which 4 ?

  21. clamin
    • one year ago
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    |dw:1441499648728:dw|

  22. clamin
    • one year ago
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    @phi

  23. phi
    • one year ago
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    -x^2 is short for -1*x*x

  24. phi
    • one year ago
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    if we use x=2 (for example) then -x^2 means -1*x*x , but with x=2, that is -1*2*2= -4 ok?

  25. clamin
    • one year ago
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    ok niw i get it..thank you

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