Correct Answer: Option D

Explanation

y = mx + c; when x = 0; y = 1

1 = m(0) + c; 1 = 0 + c; c = 1

when x = 2; y = 2

2 = m(2) + c; 2 = 2m + c; but c = 1

2 = 2m + 1

2 - 1 = 2m

2m = 1

m = \(\frac{1}{2}\)

y = \(\frac{1}{2}\)x + 1

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