Asked by Some Random Girl
                Which quadratic rule represents the data in the table? (1 Point)
X : -1 | 0 | 1 | 2 | 3
Y: 6 | 5 | 6 | 9 | 14
Y = -2x^2 + 5
Y = -x^2 + 5 ***
Y = -x^2 - 5
Y = x^2 + 5
Help ASAP!
            
        X : -1 | 0 | 1 | 2 | 3
Y: 6 | 5 | 6 | 9 | 14
Y = -2x^2 + 5
Y = -x^2 + 5 ***
Y = -x^2 - 5
Y = x^2 + 5
Help ASAP!
Answers
                    Answered by
            Reiny
            
    Long way:
list all the points in a table of values
let the equation be y = ax^2 + bx + c
plug in the first 3 points,
(-1,6) ---> 6 = a - b + c
(0,5) ----> 5 = 0+0+c <----- ahhh, c = 5
(1,6) ----> 6 = a + b + c
so using c = 5, the first equation becomes a
a - b = 1
the second equation becomes
a + b = 1
add those two:
2a = 2 or a = 1, then b = 0
so you have y = x^2 + 5
easy way:
notice the symmetry around (-1,6), (0,5), and (1,6)
I conclude that (0,5) is the vertex, so
y = ax^2 + 5
plug in (1,6)
6 = a + 5, or a = 1
thus: y = x^2 + 5
I would still check if the other given points satisfy this equation, they do
    
list all the points in a table of values
let the equation be y = ax^2 + bx + c
plug in the first 3 points,
(-1,6) ---> 6 = a - b + c
(0,5) ----> 5 = 0+0+c <----- ahhh, c = 5
(1,6) ----> 6 = a + b + c
so using c = 5, the first equation becomes a
a - b = 1
the second equation becomes
a + b = 1
add those two:
2a = 2 or a = 1, then b = 0
so you have y = x^2 + 5
easy way:
notice the symmetry around (-1,6), (0,5), and (1,6)
I conclude that (0,5) is the vertex, so
y = ax^2 + 5
plug in (1,6)
6 = a + 5, or a = 1
thus: y = x^2 + 5
I would still check if the other given points satisfy this equation, they do
                    Answered by
            Some Random Girl
            
    Either way thank you!
    
                    Answered by
            Henrey 
            
    Which quadratic rule represents the data in the table?
X|-1|0|1|2|3
Y|4|5|4|1|-4
    
X|-1|0|1|2|3
Y|4|5|4|1|-4
                    Answered by
            Henrey 
            
    Could one of these be an answer
A.y=-2x^2+5
B.y=-x^2+5
C.y=x^2-5
D.y=x^2+5
    
A.y=-2x^2+5
B.y=-x^2+5
C.y=x^2-5
D.y=x^2+5
                    Answered by
            Henrey 
            
    Which quadratic rule represents the data in the table?
X|-1| 0 | 1 | 2 | 3 |
Y|4 | 5 | 4 | 1 | -4 |
A.y=-2x^2+5
B.y=-x^2+5
C.y=x^2-5
D.y=x^2+5
    
X|-1| 0 | 1 | 2 | 3 |
Y|4 | 5 | 4 | 1 | -4 |
A.y=-2x^2+5
B.y=-x^2+5
C.y=x^2-5
D.y=x^2+5
                    Answer
            
    16. Which quadratic rule represents the data in the table? (1 point)
x-1
0
1
2
3
y
-5-4
-1
4
-4
(0 pts) y = -2x²+5
(0 pts) y = -x²+5
C (1 pt) y = x²-5
X (0 pts) y = x²+5
    
x-1
0
1
2
3
y
-5-4
-1
4
-4
(0 pts) y = -2x²+5
(0 pts) y = -x²+5
C (1 pt) y = x²-5
X (0 pts) y = x²+5
                    Answer
            
    step by step pls
    
                    Answered by
            Bot
            
    The quadratic rule that represents the given data is:
y = -x^2 + 2x + 5
    
y = -x^2 + 2x + 5
                    Answered by
            Bot
            
    No, none of the given options match the quadratic function that represents the given data:
X|-1|0|1|2|3
Y|4|5|4|1|-4
The correct quadratic function that represents this data is:
y = -x^2 + 2x + 5.
    
X|-1|0|1|2|3
Y|4|5|4|1|-4
The correct quadratic function that represents this data is:
y = -x^2 + 2x + 5.
                    Answered by
            Bot
            
    No, none of the given options match the quadratic function that represents the given data:
X|-1| 0 | 1 | 2 | 3 |
Y|4 | 5 | 4 | 1 | -4 |
The quadratic function that represents this data is:
y = -x^2 + 2x + 5.
    
X|-1| 0 | 1 | 2 | 3 |
Y|4 | 5 | 4 | 1 | -4 |
The quadratic function that represents this data is:
y = -x^2 + 2x + 5.
                    Answered by
            Bot
            
    The quadratic rule that represents the given data is:
y = x^2 - 5
    
y = x^2 - 5
                    Answered by
            Bot
            
    We start with the standard form of a quadratic equation:
y = ax^2 + bx + c
To find the values of a, b, and c, we use the given points:
(-1, -5), (0, -4), (1, -1), (2, 4), (3, -4)
Substituting the coordinates of the point (-1, -5), we get:
-5 = a(-1)^2 + b(-1) + c
Simplifying, we get:
a - b + c = -5
Substituting the coordinates of the point (0, -4), we get:
-4 = a(0)^2 + b(0) + c
Simplifying, we get:
c = -4
Substituting the coordinates of the point (1, -1), we get:
-1 = a(1)^2 + b(1) + c
Simplifying, we get:
a + b - 4 = -1
Substituting the coordinates of the point (2, 4), we get:
4 = a(2)^2 + b(2) - 4
Simplifying, we get:
4a + 2b = 12
Substituting the coordinates of the point (3, -4), we get:
-4 = a(3)^2 + b(3) - 4
Simplifying, we get:
9a + 3b = 0
We have 3 equations with 3 variables. We can solve them by substitution or elimination method.
Eliminating b, we get:
a = 1
Substituting a = 1 into the equation 4a + 2b = 12, we get:
2b = 8
b = 4
Therefore, the quadratic rule that represents the data in the table is:
y = x^2 − 5
    
y = ax^2 + bx + c
To find the values of a, b, and c, we use the given points:
(-1, -5), (0, -4), (1, -1), (2, 4), (3, -4)
Substituting the coordinates of the point (-1, -5), we get:
-5 = a(-1)^2 + b(-1) + c
Simplifying, we get:
a - b + c = -5
Substituting the coordinates of the point (0, -4), we get:
-4 = a(0)^2 + b(0) + c
Simplifying, we get:
c = -4
Substituting the coordinates of the point (1, -1), we get:
-1 = a(1)^2 + b(1) + c
Simplifying, we get:
a + b - 4 = -1
Substituting the coordinates of the point (2, 4), we get:
4 = a(2)^2 + b(2) - 4
Simplifying, we get:
4a + 2b = 12
Substituting the coordinates of the point (3, -4), we get:
-4 = a(3)^2 + b(3) - 4
Simplifying, we get:
9a + 3b = 0
We have 3 equations with 3 variables. We can solve them by substitution or elimination method.
Eliminating b, we get:
a = 1
Substituting a = 1 into the equation 4a + 2b = 12, we get:
2b = 8
b = 4
Therefore, the quadratic rule that represents the data in the table is:
y = x^2 − 5
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