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patriot [66]
4 years ago
15

For what values of k the relation R:{(k2 , 5), (3k, 6)} is not a function?

Mathematics
1 answer:
n200080 [17]4 years ago
7 0
<h3>Answer:  k = 0 or k = 3</h3>

Explanation:

If you have repeated x values, then you wont have a function. For example, the points (1,5) and (1,6) mean we don't have a function since the input x = 1 leads to multiple outputs y = 5 and y = 6 simultaneously. For a function to be possible, we must have any input lead to exactly one output only.

What we do is set the x coordinates (k^2 and 3k) equal to each other and solve for k

k^2 = 3k

k^2-3k = 0

k(k-3) = 0 .... factoring

k = 0 or k-3 = 0 .... zero product property

k = 0 or k = 3

If k = 0, then (k^2,5) becomes (0,5). Also, (3k,6) becomes (0,6). The two points (0,5) and (0,6) mean the graph fails the vertical line test.

Similarly, if k = 3, then (k^2,5) becomes (9,5) and (3k,6) = (9,6). Another vertical line test failure happens here to show we don't have a function.

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<h3>What's probability?</h3>

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