2. 6. Find the quadratic curve with a turning point (-2,3) and which passes through (-1,5) Solution: Consider the graph of y= a(x - h) 2 + k. This graph has turning point at (h,k). Since the required graph turns at (-2, 3), The graph can be written as y= a(x + 2) 2 + 3. Since it passes through (-1,5), 5= a(-1+2) 2 + 3. i.e 5= a+3 a=2 The equation of the required graph is y= 2(x +2) 2 + 3. y= 2(x 2 +4x+4) + 3. y= 2x 2 +8x+11.

3. 8. Sketch on separate diagram,the following curves. b) y = (x+4) (3-x) d) y==6x-x 2 Solution for 8b) The Given equation can be written as y= -(x+4)(x-3) i) The curve meets the x - axis at (-4,0) and (3,0) ii) The axis of symmetry about x=-½ iii) The max. point : Substitute x=-½ in the given equation y= -(x+4)(x-3), we get y= -(-½+4)(-½-3) y= 3.5 2 =12.25 iv) the given graph meets the y axis at (0,12) see graph

4. 8d) y==6x-x 2 Solution for 8d) The Given equation can be written as y= -(x)(x-6) i) The curve meets the x - axis at (0,0) and (6,0) ii) The axis of symmetry about x=3 iii) The max. point : Substitute x=3 in the given equation y= -(3)(3-6), we get y=9 The max. point at (3,9) iv) the given graph meets the y axis at (0,0) see graph

5. 8f) y=x 2 -5x-6 The Given equation can be written as y= (x+1)(x-6) i) The curve meets the x - axis at (-1,0) and (6,0) ii) The axis of symmetry about x=2.5 iii) The minimum. point : Substitute x=2.5 in the given equation y= (2.5+1)(2.5-6), we get y=-3.5 2 The minimum point at (3,9) iv) the given graph meets the y axis at (0,-6) see graph

6. Ex 3.2 10 The curve y=px 2 +4x+12 has a minimum value and cuts the x-axis at two points. Find the range of values of p Solution:1) Since the graph of y = px 2 +4x+12 has minimum value p must be>0 -----(1) ii) Since the graph cuts the x- axis at two point,its D>0 a = p, b = 4, c = 12 D = b 2 – 4ac >0 ( 4 ) 2 – 4 ( p) ( 12)> 0 16 –48p > 0 16 > 48p p< 1 / 3 -------(2) (1) &(2)  0<p<1/3