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values table f(z)=-5z+13

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

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log_□

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∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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values table f(z)=−5z+13

Solution

(−2, 23), (−1, 18), (0, 13), (1, 8), (2, 3)

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Solution steps

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Select z-values: −2, −1, 0, 1, 2

Calculate the values of the function −5z+13:

−5z+13, z=−2: 23

−5z+13, z=−1: 18

−5z+13, z=0: 13

−5z+13, z=1: 8

−5z+13, z=2: 3

Summarize in a value table:

`z`	−5`z`+13
−2	23
−1	18
0	13
1	8
2	3

Value points:

(−2, 23), (−1, 18), (0, 13), (1, 8), (2, 3)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

values table f(z)=-5z+13

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