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Exercise 1.4 · Q6

Q.From the curve y=xy=x, draw

(i) y=−xy=-x
(ii) y=2xy=2x
(iii) y=x+1y=x+1
(iv) y=12x+1y=\dfrac12x+1
(v) 2x+y+3=02x+y+3=0.
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Step 1 (i) y=−xy=-x. Reflection of y=xy=x about the xx-axis (equivalently about the yy-axis, since both give the same line here) -- slope flips from 11 to −1-1.

Step 2 (ii) y=2xy=2x. Vertical dilation of y=xy=x by factor 2 -- steeper line through the origin, slope 22.

Step 3 (iii) y=x+1y=x+1. Vertical shift of y=xy=x up by 1 -- parallel line through (0,1)(0,1) instead of the origin.

Step 4 (iv) y=12x+1y=\tfrac12x+1. Vertical dilation by factor 12\tfrac12 (compress toward the xx-axis, giving slope 12\tfrac12), then shift up by 1 -- line through (0,1)(0,1) with a gentler slope. …

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