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Complete the ordered pairs that lie on the graph of the function [tex]h[/tex].

[tex]
h(x)=\left\{\begin{array}{ll}
-x^2 - 6x - 9, & x \ \textless \ -2 \\
\left(\frac{1}{3}\right)^x - 4, & -2 \leq x \leq 2 \\
\frac{1}{2} x - 4, & x \ \textgreater \ 2
\end{array}\right.
[/tex]

Type the correct answer in each box. Use numerals instead of words.

(-4, [tex]\square[/tex])

(0, [tex]\square[/tex])

(8, [tex]\square[/tex])


Sagot :

We need to find the values of [tex]\(h(x)\)[/tex] for the given [tex]\(x\)[/tex]-coordinates and complete the ordered pairs.

1. For [tex]\( x = -4 \)[/tex]:
Since [tex]\( -4 < -2 \)[/tex], we use the first piece of the function:
[tex]\[ h(x) = -x^2 - 6x - 9 \][/tex]
Plugging in [tex]\( x = -4 \)[/tex]:
[tex]\[ h(-4) = -(-4)^2 - 6(-4) - 9 \][/tex]
[tex]\[ h(-4) = -16 + 24 - 9 \][/tex]
[tex]\[ h(-4) = -1 \][/tex]

So, the ordered pair is [tex]\( (-4, -1) \)[/tex].

2. For [tex]\( x = 0 \)[/tex]:
Since [tex]\(-2 \leq 0 \leq 2\)[/tex], we use the second piece of the function:
[tex]\[ h(x) = \left( \frac{1}{3} \right)^x - 4 \][/tex]
Plugging in [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = \left( \frac{1}{3} \right)^0 - 4 \][/tex]
[tex]\[ h(0) = 1 - 4 \][/tex]
[tex]\[ h(0) = -3 \][/tex]

So, the ordered pair is [tex]\( (0, -3) \)[/tex].

3. For [tex]\( x = 8 \)[/tex]:
Since [tex]\( 8 > 2 \)[/tex], we use the third piece of the function:
[tex]\[ h(x) = \frac{1}{2}x - 4 \][/tex]
Plugging in [tex]\( x = 8 \)[/tex]:
[tex]\[ h(8) = \frac{1}{2}(8) - 4 \][/tex]
[tex]\[ h(8) = 4 - 4 \][/tex]
[tex]\[ h(8) = 0 \][/tex]

So, the ordered pair is [tex]\( (8, 0) \)[/tex].

Thus, completing the ordered pairs:
[tex]\[ (-4, \mathbf{-1}),\ (0, \mathbf{-3}),\ (8, \mathbf{0}) \][/tex]