1. anonymous

Okay let's solve for x first. Have you heard of the quadratic formula?

2. clamin

Yes. y=ax^2+bx+c @sepeario

3. 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 :)

4. clamin

This helps a little but I'm looking for the x and y value which I struggle a lot to find..@leong

5. clamin

@Leong

6. anonymous

huh?

7. 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.....

8. 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 :)

9. clamin

Its for the graph

10. freckles

what do you mean to show you the x and y value?

11. freckles

x and y are variables they vary they don't just take on one value

12. anonymous

i think he/she means solve for x and y.

13. Nnesha

x and y intercepts ? @clamin ?

14. 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

15. 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

16. 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

17. 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)

18. clamin

@phi how come the 4 is negative?

19. clamin

@nnesha @freckles @sepeario @leong

20. phi

which 4 ?

21. clamin

|dw:1441499648728:dw|

22. clamin

@phi

23. phi

-x^2 is short for -1*x*x

24. 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?

25. clamin

ok niw i get it..thank you