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=2The equation of the required graph is

y= 2(x +2)2 + 3.

y= 2(x 2+4x+4) + 3.

y= 2x 2+8x+11.

8. Sketch on separate diagram,the following curves.

b) y = (x+4) (3-x)

d) y==6x-x2

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= 3.52 =12.25

iv) the given graph meets the y axis at (0,12) see graph

8d) y==6x-x2

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

8f) y=x2-5x-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=-3.52

The minimum point at (3,9)

iv) the given graph meets the y axis at (0,-6) see graph

Ex 3.2 10 The curve y=px2+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 = px2+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 = b2 – 4ac >0

( 4 )2 – 4 ( p) ( 12) > 0

16 –48p > 0

16 > 48p

p< 1/3 -------(2)

(1) &(2) 0<p<1/3