For questions 1 and 2, use integer values of x from –3 to 3 to graph the equation.
1. y = –2x2 + 3 (1 point)
To graph the equation y = -2x^2 + 3, we need to substitute different integer values of x from -3 to 3 and calculate the corresponding values of y.
Let's start by using x = -3:
y = -2(-3)^2 + 3
y = -2(9) + 3
y = -18 + 3
y = -15
So, the first point on the graph is (-3, -15).
Next, let's use x = -2:
y = -2(-2)^2 + 3
y = -2(4) + 3
y = -8 + 3
y = -5
The second point on the graph is (-2, -5).
By continuing this process for x = -1, 0, 1, 2, and 3, we can find the rest of the points:
For x = -1: y = -2(-1)^2 + 3 = -2 + 3 = 1
Point: (-1, 1)
For x = 0: y = -2(0)^2 + 3 = 0 + 3 = 3
Point: (0, 3)
For x = 1: y = -2(1)^2 + 3 = -2 + 3 = 1
Point: (1, 1)
For x = 2: y = -2(2)^2 + 3 = -8 + 3 = -5
Point: (2, -5)
For x = 3: y = -2(3)^2 + 3 = -18 + 3 = -15
Point: (3, -15)
Now we have the coordinates of all the points. We can plot them on a graph and then connect them to get the graph of the equation y = -2x^2 + 3.
Use the following points for graphing:
(X,Y)
(-3,-15)
(-2,-5)
(-1,1)
V(0,3)
(1,1)
(2,-5)
(3,-15).
NO HELP IDIOTS
henry could you please give te answers
1. A
2. D
3. D
4. D
5. B
6. D
7. D
100
thanks hal
is Hal correct
Hal is correct and just to be nice i know this can be hard
For the essays put this is your own words
A) The coordinates would be A(1,-3) B(4,-5) C(3,-1)
B) The x-coordinate stays the same and the y-coordinates is the same distance from the x-axis but on the opposite side of the x axis, the negative side.
C) The new coordinates are A(2,1) B=(5,3) C(4,-1)
D) Translation arrow notation; (x,y)?(x+1,y-2)