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Exercise 1.4 · Q1

Q.For the curve y=x3y=x^3, draw

(i) y=−x3y=-x^3
(ii) y=x3+1y=x^3+1
(iii) y=x3−1y=x^3-1
(iv) y=(x+1)3y=(x+1)^3 with the same scale.
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Step 1 (i) y=−x3y=-x^3. This is y=−f(x)y=-f(x) for f(x)=x3f(x)=x^3 -- a reflection about the xx-axis. Since x3x^3 is an odd function, −x3=(−x)3-x^3=(-x)^3, so this reflection actually looks identical in shape to the ORIGINAL curve rotated -- but formally it is the xx-axis mirror image. Key points: (1,1)→(1,−1)(1,1)\to(1,-1), (−1,−1)→(−1,1)(-1,-1)\to(-1,1), centre point (0,0)(0,0) unchanged.

Step 2 (ii) y=x3+1y=x^3+1. This is y=f(x)+1y=f(x)+1 -- shift the whole curve UP by 1 unit. Key points: (0,0)→(0,1)(0,0)\to(0,1), (1,1)→(1,2)(1,1)\to(1,2), (−1,−1)→(−1,0)(-1,-1)\to(-1,0).

Step 3 (iii) y=x3−1y=x^3-1. Shift DOWN by 1 unit. Key points: (0,0)→(0,−1)(0,0)\to(0,-1), (1,1)→(1,0)(1,1)\to(1,0), (−1,−1)→(−1,−2)(-1,-1)\to(-1,-2).

Step 4 (iv) y=(x+1)3y=(x+1)^3. This is y=f(x+1)y=f(x+1) -- shift LEFT by 1 unit (since c=1>0c=1>0 in f(x+c)f(x+c)). Key points: (0,0)→(−1,0)(0,0)\to(-1,0), (1,1)→(0,1)(1,1)\to(0,1), (−1,−1)→(−2,−1)(-1,-1)\to(-2,-1).

✓Final answer

All four are drawn on y=x3y=x^3's frame: (i) reflect about xx-axis;

(ii) shift up 1;

(iii) shift down 1;

(iv) shift left 1 -- each keeping the same S-shape as y=x3y=x^3, just moved/mirrored per the rule above.

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