Write a quadratic function rule for the given table in problems 4 and 5.
4. X (0, 1, 2, 3)
Y (-6, -5, -7, 3)
a. y= x-6
b. y= x-4
c. y= x^2 - 3
d. y= x^2 - 6
5. X(-1, 0, 1, 7)
Y(5,4,5,8)
a.y=x+4
b.y=x^2+4
c.y= x^2+6
d.y=x+6
#4 cannot be correct
y increases, then decreases, then increases again.
However, if y(2) = -2, y = x^2-6
For #5, clearly (a) and (b) are out.
Since 5 = 1+4, (c) must be correct, but there is a other huge typo. y(2) = 8, NOT y(7)
C
A
D
D
C
^^ answers. 100%
No it's
C
A
D
D
B
These are for all connections student good luck
thank you bea miller :)
sam you had a typo in .4 you wrote -7 it was -2 and bea miller is correct 100%
To find the quadratic function rule that represents the given table, we need to analyze the pattern in the y-values for different x-values.
For problem 4:
Let's examine the differences between consecutive y-values:
-5 - (-6) = 1
-7 - (-5) = -2
3 - (-7) = 10
As you can see, the differences are not constant, which indicates that the relationship between x and y is non-linear. We can eliminate options a and b.
Now let's test options c and d by substituting the x-values into the functions and comparing the results to the actual y-values:
Option c: y = x^2 - 3
For x = 0: 0^2 - 3 = -3 (Incorrect, actual y-value is -6)
For x = 1: 1^2 - 3 = -2 (Incorrect, actual y-value is -5)
For x = 2: 2^2 - 3 = 1 (Incorrect, actual y-value is -7)
For x = 3: 3^2 - 3 = 6 (Incorrect, actual y-value is 3)
Option d: y = x^2 - 6
For x = 0: 0^2 - 6 = -6 (Correct, actual y-value is -6)
For x = 1: 1^2 - 6 = -5 (Correct, actual y-value is -5)
For x = 2: 2^2 - 6 = -2 (Incorrect, actual y-value is -7)
For x = 3: 3^2 - 6 = 3 (Correct, actual y-value is 3)
Out of options c and d, option d yields more correct results. Therefore, the quadratic function rule for problem 4 is:
d. y = x^2 - 6
For problem 5:
Similarly, let's examine the differences between consecutive y-values:
4 - 5 = -1
5 - 4 = 1
8 - 5 = 3
The differences are not constant, so we can eliminate options a and d.
Now let's test options b and c:
Option b: y = x^2 + 4
For x = -1: (-1)^2 + 4 = 1 + 4 = 5 (Correct, actual y-value is 5)
For x = 0: (0)^2 + 4 = 0 + 4 = 4 (Correct, actual y-value is 4)
For x = 1: (1)^2 + 4 = 1 + 4 = 5 (Correct, actual y-value is 5)
For x = 7: (7)^2 + 4 = 49 + 4 = 53 (Incorrect, actual y-value is 8)
Option c: y = x^2 + 6
For x = -1: (-1)^2 + 6 = 1 + 6 = 7 (Incorrect, actual y-value is 5)
For x = 0: (0)^2 + 6 = 0 + 6 = 6 (Incorrect, actual y-value is 4)
For x = 1: (1)^2 + 6 = 1 + 6 = 7 (Incorrect, actual y-value is 5)
For x = 7: (7)^2 + 6 = 49 + 6 = 55 (Incorrect, actual y-value is 8)
Out of options b and c, option b yields more correct results. Therefore, the quadratic function rule for problem 5 is:
b. y = x^2 + 4