Question

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 = (

Answer 1

HELP PLEASE

Answer 2

what you want to do?

Answer 3

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

Answer 4

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)

Answer 5

I got the answer a is that wrong

Answer 6

sorry ali, but i dont understand what you want

Answer 7

Functions and Polynomials

Answer 8

is the subject

Answer 9

if you cant help i understand

Answer 10

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)

Answer 11

If you still need it.

Answer 12

@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