anonymous
  • anonymous
y = –x2 + 2 3. When x = –2 (1 point) y = –(–2)2 + 2 = –2; (–2, –2) y = (–2)2 + 2 = 2; (–2, 2) y = –(2)2 + 2 = –2; (2, –2) y = –(–2)2 – 2 = 2; (–2, –2) 4. When x = – 1 (1 point) y = –(–1)2 + 1 = 2; (–1, 1) y = (–1)2 + 2 = 1; (–2, 1) y = –(–1)2 + 2 = 1; (–1, 1) y = (–1)2 + 2 = 1; (1, –1) 5. When x = 0 (1 point) y = –(0)2 + 2 = 2; (0, 2) y = –(0)2 + 2 = –2; (0, –2) y = (0)2 – 2 = 2; (2, 0) 6. When x = 1 (1 point) y = –(1)2 – 2 = 1; (1, 1) y = –(1)2 + 2 = 1; (1, 1) y = (1)2 + 2 = –1; (1, 1) y = (
Pre-Algebra
chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
HELP PLEASE
anonymous
  • anonymous
what you want to do?

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anonymous
  • anonymous
I need to Complete the table and choose the solution for each given value of x in questions 3–7. And I cant get it its frutsrating
anonymous
  • anonymous
If you want to fill the table just do what it says on the header. For the first line use x= -2 and calculate the Y. When x = –2 (1 point) y = –(–2)2 + 2 = –2; y = (–2)2 + 2 = 2; y = –(2)2 + 2 = –2; y = –(–2)2 – 2 = 2; (–2, –2)
anonymous
  • anonymous
I got the answer a is that wrong
anonymous
  • anonymous
sorry ali, but i dont understand what you want
anonymous
  • anonymous
Functions and Polynomials
anonymous
  • anonymous
is the subject
anonymous
  • anonymous
if you cant help i understand
anonymous
  • anonymous
x=-2 y=-2 (-2;-2) x=-1 y= 1 (-1; 1) x= 0 y= 2 (0;2) x= 1 y= 1 (1;1) x= 2 y= -2 (2;-2)
anonymous
  • anonymous
If you still need it.
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anonymous
  • anonymous
@LILKITTYNICKLE that's wrong. You messed all the minus signals. for the first example where x= -2 you can't simply cut this minus with the minus on the formula. it'll be: \[y= -x ^{2}+2 => y = - (-2)^{2} +2 => y = - 4 + 2 = -2\] y= -2, not 6

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