Mathematics, 22.01.2021 09:00 jeanniebyrd54
6. Prove that linear functions grow by equal differences over equal intervals. Part I. This is the graph of . Use the graph to show that equal intervals of x-values have equal differences of y-values. a) Think about an interval on the x-axis starting with p and ending with p + k. What is the difference between the x-values? What is the difference between the y-values for these x-values? Complete the tables using . Interval between p = 2 and p + k = 6 Difference x-value 2 6 4 y-value ā2 Interval between p = 10 and p + k = 14 Difference x-value 10 14 y-value b) The difference in the x-values for each interval is . This value is constant because difference = (p + k) ā p = . c) The difference in the y-values for each interval is . This value is also constant because \begin{aligned} \text{difference } &= -\frac{1}{2}(p + k) + 10 - (-\frac{1}{2}p + 10) \\ &= -\frac{1}{2}p - \frac{1}{2}k + 10 + \frac{1}{2}p - 10 \\ &= -\frac{1}{2}k \end{aligned} difference =ā 2 1 (p+k)+10ā(ā 2 1 p+10) =ā 2 1 pā 2 1 k+10+ 2 1 pā10 =ā 2 1 k d) Does your answer in part c depend on p? If not, what does it depend on? Part II. Use a similar method to prove that for any linear equation, y = mx + b, equal intervals result in growth by equal differences. Step 1: If the equal interval length is k, the two endpoints of the interval will be labeled p and p + k. Label the endpoints of the corresponding interval on the y-axis of the graph below. Hint: First, substitute x = p into y = mx + b. Next, substitute x = p + k into y = mx + b. Step 2: Find the difference between the endpoints of the interval on the y-axis. (1) (2) [y-value at x = p] ā [y-value at x = p + k] Step 3: Interpret your answer from Step 2 to prove that linear functions always grow by equal differences over equal intervals. Hint: Use these questions to justify your Does your answer to Step 2 depend on p? What does it depend on? Is the difference between the y-values the same regardless of where the inter
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6. Prove that linear functions grow by equal differences over equal intervals. Part I. This is the g...
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