Answer:
A. (3,-4) fall on the graph of f(x) = -3 x² + 6x + 5.
B. (-1,4) DO NOT fall on the graph of f(x) = -3 x² + 6x + 5.
Step-by-step explanation:
Here, the given function is: f(x) = -3 x² + 6 x + 5
Now, as we know f(x) = y
⇒ y = -3 x² + 6x + 5
Now, to show if any arbitrary point (a,b) is on the graph of y,
we need to show that for y = b, and x = a the Left side of expression =Right side of expression.
A. Consider the point (3,-4)
Here, y = -4, and x = 3
Solving -3 x² + 6 x + 5 for x = 3, we get:
-3 (3)² + 6 (3) + 5 = -27 + 18 + 5 = -27 + 23 = -4 = y
⇒f(3) = y = -4 ⇒ Left Side = Right side
Hence, (3,-4) fall on the graph of f(x) = -3 x² + 6x + 5.
A. Consider the point (-1,4)
Here, y = 4, and x = -1
Solving -3 x² + 6 x + 5 for x = -1, we get:
-3 (-1)² + 6 (-1) + 5 = -3 -6 + 5 = -4 ≠ y
⇒f(-1) = y ≠ 4 ⇒ Left Side≠ Right side
Hence, (-1,4) DO NOT fall on the graph of f(x) = -3 x² + 6x + 5.
