Qualification > Math
Math paper4 Jun. 2009 help
moon:
I need help in NOV 2004 Q7 Please.It is diffiicult.
Ghost Of Highbury:
--- Quote from: moon on November 03, 2009, 03:58:54 pm ---I need help in NOV 2004 Q7 Please.It is diffiicult.
--- End quote ---
the whole q?
moon:
please I need a3, bii and c2 ???
wassup:
--- Quote from: moon on November 03, 2009, 03:58:54 pm ---I need help in NOV 2004 Q7 Please.It is diffiicult.
--- End quote ---
7 a)i) y = px + qx + r
y = 4^2+(-2*4)+(-3)
y = 5
ii) On K and L y=0
so, px + qx + r=0
x-2x-3=0
Solve this n u'll get x=-1, 3
K(-1,0) L(3,0)
iii) On M x=0
so, y=r
y=-3
M(0,-3)
b)i) Upside down
ii) Eqn, will be x^2=y, so just put some values n get the shape of graph
c)i) c=0 because it passes through d origin
ii) (3,0) (4,8)
put these in the eqn.
3^2a + 3b = 0
4^2a + 4b = 8
solve these n u'll get a = 2, b = –6
Ghost Of Highbury:
--- Quote from: moon on November 03, 2009, 04:17:23 pm ---please I need a3, bii and c2 ???
--- End quote ---
for aiii
best wud be to draw the graph, and see where the minimum point lies. write its cordinates
for confirmation u can use differentiation
dy/dx = 2x - 2
2x - 2 =0
2x = 2
x = 1 substituting y = 4
dont bother if u dont understand this
for bii
u know then the value are applied in the equation
u r left with only y = x2 which is a parabola and passes through the origin (0,0)
for cii
it says it passes through the origin
which means, c =0
u r left with
y= ax^2 + bx
substitute the cordinates (3, 0) and (4, 8).
to get
0 = 9a + 3b
and
8 = 16a + 4b
from the second we can get
a = 8-4b / 16
substitute this in the first to get
72 - 36b
-------- = -3b
16
72 - 36b = -48b
solve to get
b = -6 and a = 2
@wassup - for aiii - it is not they-axis, its a line of symmetry
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