The domain are the possible input while the range are the possible output
of a function.
Reasons:
The given functions can be expressed by the equation; (-x + 1)·(x + 1) = -x² + 1
Therefore, we have;
(a) y = f(x) + 1 = -x² + 1 + 1 = -x² + 2
The x-intercept of the above function are, x = √2, and x = -√2
Which gives;
The domain = [-√2, √2]
The range = [0, 2]
(b) y = 3·f(x) = 3 × (-x² + 1) = -3·x² + 3
At the x–intercepts, we have;
-3·x² + 3 = 0
x = ±1
The domain = [-1, 1]
The maximum value of y is given at x = 0, therefore;
= -3 × 0² + 3 = 3
The range = [0, 1]
(c) y = -f(x) = -(-x² + 1) = x² - 1
At the x–intercepts, x² - 1 = 0
x = ± 1
The domain = [-1, 1]
The minimum value of y is given at x = 0, which is y = -1
The range = [0, -1]
(d) y = f(x - 1) = -(x - 1)² + 1 = -x² + 2·x
At the x–intercepts, we have; -x² + 2·x = 0, which gives;
(-x + 2)·x = 0
Which gives, x = 0, or x = 2
The domain = [0, 2]
The maximum value of y is given when x = -b/(2·a) = -2/(2×(-1)) = 1
y = f(1) = -1² + 2×1 = 1
Therefore;
The range = [0, 1]
(e) y = f(x + 2) + 1 = (-(x + 2)² + 1) + 1 = -x² - 4·x - 2
At the x–intercepts, we have; -x² - 4·x - 2 = 0, which gives;
x = -(2 + √2) or x = x = √2 - 2
The domain = [-(2 + √2), (√2 - 2)]
The maximum value of y is given when x = -4/(2)) = -2
Which gives;
-(-2)² - 4·(-2) - 2 = 2
The range = [0, 2]
Learn more here:
https://brainly.com/question/20391090
