Question

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

Answer 1

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

Answer 2

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

Answer 3

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 :)

Answer 4

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

Answer 5

@Leong

Answer 6

huh?

Answer 7

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.....

Answer 8

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 :)

Answer 9

Its for the graph

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

x and y intercepts ? @clamin ?

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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)

Answer 18

@phi how come the 4 is negative?

Answer 19

@nnesha @freckles @sepeario @leong

Answer 20

which 4 ?