5.Curves Sketching Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Curve Sketching T1 Solutions Curve Sketching 1(i) Asymptotes: 7yx and 2x (ii) 5 2 2 5 2 2k    (iii) 2(i) Method 1: Using Differentiation 22 11 (2 ) dy dx a x x   Putting 0,dy dx  22 11 (2 )a x x   22(2 ) (2 )a x x x a x     xa 1 1 2 2y a a a a   2 2 3 3 22 (2 ) dy dx a x x   When xa , 2 2 3 3 22 0dy dx a a    2,a a   is a min point. Method 2: Non-differentiation method 1 1 2 2 (2 ) ay a x x x a x   is the reciprocal graph of quadratic eqn 𝑦 = 𝑥(2𝑎 − 𝑥) and multiplied by scale factor 2a in the y-direction. 7yx 2x  3.41, 2.17  0.586,7.83

River Valley High School, Mathematics Department 2023 Curve Sketching T2 Since the graph of 𝑦 = 𝑥(2𝑎 − 𝑥) is a parabola with a max point at (𝑎 , 𝑎2), the curve 𝐶1 will have a min point at 𝑥 = 𝑎, and 𝑦 = 2𝑎 ( 1 𝑎2) = 2 𝑎. Thus 2,a a   is a min point. (ii) (iii)   2 2 22 1xa y ab   . The graph of 𝐶2 is an ellipse with centre (𝑎, 0), with 2 end-pts at (0,0) and (2a,0). For the ellipse to intersect 𝐶1 at more than one pt, 𝑏 > 2 𝑎. (iv) 2a and 6b 2 2 1 1 ( ) 12 xaba x x a         From the graph, 0 < 𝑥 < 0.330 or 3.67 < 𝑥 < 4 (3 s.f.)

River Valley High School, Mathematics Department 2023 Curve Sketching T3 3 (i) Since C has a vertical asymptote at 0x , thus 1 0, 1    . (ii) 22 4 2 4 3 3 3 xyx xx    ; Oblique asymptote is 2 3yx (iii) 224 3 xy x  23 2 4xy x 22 3 4 0x xy   The equation above has no real roots when discriminant < 0.      2 2 3 4 2 4 0 9 32 0 32 32( )( ) 033 y y yy        M1, for getting b2-4ac <0 correctly 4 2 4 2 33 y   2a . (iv) 22 4 2 2 6 3( 1) 3 3 3( 1) xy k x kxx       (a) x – intercepts are 23 9 3 24 16 2 k k k    and 23 9 3 24 16 2 k k k    4/3 +k -1

River Valley High School, Mathematics Department 2023 Curve Sketching T4 (b) For the graph of 224 3( 1) xyk x   to cut the x-axis at 2 distinct points, 293 216 2 kk  > 0, =>least integer value of k is 4 4 5x is an asymptote 5d    2 555 ax bx c Ry ax b axx      7yx is an asymptote 1, 5 7 2 a b a b             22 2 5 2 2 22d 5d 5 x x x x cx x c yy xx x         At 1x ,     d 0 1 5 2 2 1 2 0 1d y ccx             Sketch 2 21 5 xxy x       21 2 5 1 0xx     22 2 5 1 0x x x x      22 2 1 5x x x x      2 21 5 xx xx   Therefore, sketch the graph of yx  on the same diagram. (i) yx  , 1  From graph, only one pt of intersection  one real root

River Valley High School, Mathematics Department 2023 Curve Sketching T5 (ii) yx  ,