place each of the key aspects under the function it describes
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Answer:
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)
Step-by-step 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 , ∞)
First function h(x) is given by:
[tex]h(x)=(x-1)^2[/tex]
The function is a quadratic function
and also we know that:
[tex](x-1)^2\geq 0[/tex]
Hence, the range is: {y | 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:
[tex]g(x)=-(x+1)(x-1)[/tex]
which could also be written by:
[tex]g(x)=-(x^2-1)[/tex]
since we know that:
[tex](a+b)(a-b)=a^2-b^2[/tex]
Also,we know that:
[tex]x^2\geq 0\\\\x^2-1\geq -1\\\\-(x^2-1)\leq 1[/tex]
i.e.
[tex]g(x)\leq 1[/tex]
Hence, we get the range of this function is: {y | 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,∞).