Question

Place each of the key aspects under the function it describes

Answer 1

Answer with explanation:

First function h(x) is given by:

`h(x)=(x-1)^2`

The function is a quadratic function

and also we know that:

`(x-1)^2\geq 0`

Hence, the range is: y≥0

Also, we know that the graph of the function decreases continuously in the interval (-∞,1) and then it increases continuously in the interval (1,∞)

and the vertex is at (1,0)

Second function is:

`g(x)=-(x+1)(x-1)`

which could also be written by:

`g(x)=-(x^2-1)`

since we know that:

`(a+b)(a-b)=a^2-b^2`

Also,we know that:

`x^2\geq 0\\\\x^2-1\geq -1\\\\-(x^2-1)\leq 1`

i.e.

`g(x)\leq 1`

Hence, we get the range of this function is: y≤1

The graph of this function is a downward open parabola with vertex at (0,1) and hence, the graph is increasing in the interval (-∞,0) and decreasing over the interval (0,∞).

Answer 2

range: {y: y ≥ 0} ⇒ h(x)

decreasing over(-∞ , 1) ⇒ h(x)

increasing over (-∞ , 0) ⇒ g(x)

range: {y: y ≤ 1} ⇒ g(x)

increasing over (1 , ∞) ⇒ h(x)

decreasing over(0 , ∞) ⇒ g(x)

Explanation:

∵ h(x) = (x - 1)²

∴ Its vertex is (1 , 0)

∴ The parabola is opened upward

∴ The range: {y: y ≥ 0}

∴ It's decreasing over (-∞ , 1)

∴ It's increasing over (1 , ∞)

∵ g(x) = -(x - 1)(x + 1) = -(x² - 1) = -x² + 1

∴ Its vertex is (0 , 1)

∴ The parabola is opened downward

∴ The range: {y: y ≤ 1}

∴ It's increasing over (-∞ , 0)

∴ It's decreasing over (0 , ∞)