clamin
  • clamin
PLEASE HELP!!! MEDAL!!! Please show me the x and y value of y=-x^2+2x-2
Mathematics
jamiebookeater
  • jamiebookeater
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Sepeario
  • Sepeario
Okay let's solve for x first. Have you heard of the quadratic formula?
clamin
  • clamin
Yes. y=ax^2+bx+c @sepeario
anonymous
  • anonymous
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 :)

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clamin
  • clamin
This helps a little but I'm looking for the x and y value which I struggle a lot to find..@leong
clamin
  • clamin
anonymous
  • anonymous
huh?
anonymous
  • anonymous
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.....
anonymous
  • anonymous
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 :)
clamin
  • clamin
Its for the graph
freckles
  • freckles
what do you mean to show you the x and y value?
freckles
  • freckles
x and y are variables they vary they don't just take on one value
Sepeario
  • Sepeario
i think he/she means solve for x and y.
Nnesha
  • Nnesha
x and y intercepts ? @clamin ?
Nnesha
  • Nnesha
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
phi
  • phi
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
phi
  • phi
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
phi
  • phi
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)
clamin
  • clamin
@phi how come the 4 is negative?
clamin
  • clamin
phi
  • phi
which 4 ?
clamin
  • clamin
|dw:1441499648728:dw|
clamin
  • clamin
phi
  • phi
-x^2 is short for -1*x*x
phi
  • phi
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?
clamin
  • clamin
ok niw i get it..thank you

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