The figure below shows the graph of function f with domain [-1,1] and range [0,1].

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]

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https://brainly.com/question/20391090

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