SAJC Check your Understanding_Graphing_teacher

1 Check your Understanding (Graphing) Section 1:Asymptotes MUTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1. What is the vertical asymptote(s) of 1h( ) 4x x= − ? (a) None (b) 4x= (c) 4x=− (d) 1x= (b) 2. What is the horizontal asymptote(s) of 1h( ) 2 4x x=+ − ? (a) None (b) 4y= (c) 2y= (d) 4x= (c) 3. What is the vertical asymptote(s) of h( ) 4 xx x= − ? (a) None (b) 4x= (c) 4x=− (d) 1x= (b) 4. What is the horizontal asymptote(s) of h( ) 4 xx x= − ? (a) None (b) 4y= (c) 2y= (d) 1y= (d)

2 5. What is the horizontal asymptote(s) of 2 2 32h( ) 85 xx x= − ? (a) None (b) 4y= (c) 5y= (d) 5y= (b) 6. What is the vertical asymptote(s) of ( )( ) 5h( ) 98 xx xx −= −+ ? (a) 9, 8xx= =− (b) 5x= (c) 9, 8xx=− = (d) 5x=− (a) 7. What is the horizontal asymptote(s) of 2 34h( ) 4 xxx x +−= − ? (a) None (b) 4y= (c) 2y= (d) 3y=− (d)

3 Match the equation with the appropriate graph? 8. 2 2 2() 4 xfx x= − ( D )

4 Section 2:Stationary points 9. RI Promo 8865/2018/Q6 (part) The curve C has equation y = f (x) where f (x) = 6 x − 3x 2 − 4 x 3. (i) Find dy dx . Hence find the coordinates of the stationary points on the curve. (ii) Use a non-calculator method to determine the nature of each of the stationary points. Solution: (i) dy dx = 6 − 6x − 12x 2 For stationary points, dy dx = 0  6 − 6x − 12x 2 = 0  x = −1 or 1 2 When x = −1, y = −5 When x = 1 2 , y = 7 4 Coordinates of the stationary points are (−1, −5) and   1 2  7 4 . (ii) d 2 y dx 2 = −6 − 24x When x = −1, d 2 y dx 2 = 18  0 (minimum point) When x = 1 2 , d 2 y dx 2 = −18  0 (maximum point)  (−1, −5) is a minimum point and   1 2  7 4 is a maximum point.

5 Section 3: Graphing Problems 10. RIJC Promo 8865/2015/Q2 Sketch the curve with equation y = x + 2 x + 1 , stating the equations of any asymptotes and the coordinates of the points where the curve crosses the axes. Solution: y = x + 2 x + 1 y x (−2, 0) O (0, 2) y = 1 x = −1

6 11. MIJC Promo 8865/2015/Q5 (i) Sketch the graph of y = 1 4x− . [2] (ii) By drawing an additional graph in (i), solve the inequality 1 4x− – 3 ≤ 0. [2] (iii) Hence solve the inequality 1 e4x− ≤ 3. [3] Answer: (ii) 4x or 4.33x . (3 s.f.) (iii) ln 4x or 1.47x . (3 s.f.) Solution: 11(i) (ii) 1 4x− – 3 ≤ 0  1 4x− ≤ 3 From graph, 4x or 4.33x . (3 s.f.) (iii) 1 e4x− ≤ 3  1 e4x− – 3 ≤ 0  e4x  or e 4.3333x   ln 4x or 1.47x . (3 s.f.) x y = 1 4x− x = 4