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clamin
 one year ago
PLEASE HELP!!! MEDAL!!!
Please show me the x and y value of y=x^2+2x2
clamin
 one year ago
PLEASE HELP!!! MEDAL!!! Please show me the x and y value of y=x^2+2x2

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay let's solve for x first. Have you heard of the quadratic formula?

clamin
 one year ago
Best ResponseYou've already chosen the best response.0Yes. y=ax^2+bx+c @sepeario

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you 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+2x2 : 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: \[y= \frac{ 2\pm \sqrt{2^{2}4(1)(2)} }{ 2(1) }\] \[y=\frac{ 2\pm \sqrt{48} }{ 2 }\] Since \[\sqrt{48} = \sqrt{4}\] which doesn't make sense, so it will be false! which now we know that it will be NO SOLUTION! :) hope it help :)

clamin
 one year ago
Best ResponseYou've already chosen the best response.0This helps a little but I'm looking for the x and y value which I struggle a lot to find..@leong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you 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
 one year ago
Best ResponseYou've already chosen the best response.0not 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 :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean to show you the x and y value?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0x and y are variables they vary they don't just take on one value

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think he/she means solve for x and y.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0x and y intercepts ? @clamin ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0if yes then use \(\huge\color{reD}{\rm b^24ac}\) `Discriminant` if ` b^24ac > 0` then there are 2 real zeros if ` b^24ac = 0` then there is one real root if ` b^24ac < 0` then you will get two complex roots (no xintercept) to find yintercept substitute x for 0 solve for y

phi
 one year ago
Best ResponseYou've already chosen the best response.1If 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
 one year ago
Best ResponseYou've already chosen the best response.1let'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
 one year ago
Best ResponseYou've already chosen the best response.1now 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+42 = 2 (2,2) x=3 y= (3*3) +2*3 2 = 9 +62 = 5 (3,5)

clamin
 one year ago
Best ResponseYou've already chosen the best response.0@phi how come the 4 is negative?

clamin
 one year ago
Best ResponseYou've already chosen the best response.0@nnesha @freckles @sepeario @leong

phi
 one year ago
Best ResponseYou've already chosen the best response.1if 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
 one year ago
Best ResponseYou've already chosen the best response.0ok niw i get it..thank you
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