Answer:
B, A, B.
Step-by-step explanation:
Using a substitution method, it can be shown which (x,y) point does not satisfy the given equations.
1. Starting from the first function, represented algebraically by the equation 3x^2+6-4:
f(x)= -3x^2+6x-4
f(6)= -3(6)^2+6(6)-4
f(6)= (-3)(36)+36-4
f(6)= -76
That's why the point (x,y)= (6,-70) is not on the graph of y=-3x^2+6x-4
The point that is on the graph is (x,y)= (6,-76)
The answer is B. (6,-70)
2. For the second function, represented algebraically by the equation 2x^2+3x-10:
f(x)= 2x^2+3x−10
f(4)= 2(4)^2+3(4)-10
f(4)= 32+12-10
f(4)= 34
That's why the point (x,y)= (4,30) is not on the graph of y= 2x^2+3x−10
The point that is on the graph is (x,y)= (4,34)
The answer is A. (4,30)
3. For the last function, represented algebraically by the equation 1/2x^2-3x+7:
f(x)= −1/2 x^2 −3x+7
f(6)= −1/2 (6)^2 −3(6)+7
f(6)= −1/2 (36)-18+7
f(6)= -18-18+7
f(6)= -29
That's why the point (x,y)= (6,-30) is not on the graph of y= −1/2 x^2 −3x+7
The point that is on the graph is (x,y)= (6,-29)
The answer is B. (6,-30)
