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secondary 3 | A Maths
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Date Posted: 4 years ago
Views: 398
(i)
k = ½
bcos x = 0, y = ½
(ii)
eˣ + x/k - 2/k = 0
re-arranging, ...
keˣ = (2 - x)
this is equivalent to simultaneous soln of the following eqns:
y = keˣ . . . . . . . . . [1]
y = 2 - x . . . . . . . . . [2]
if we draw the graph of y = 2-x on the graph of y = keˣ ,
we shld obtain 1 intersection pt.
hence, one solution to eˣ + x/k - 2/k = 0
(iii)
to sketch y = keˣ + 4,
draw the same graph of y = keˣ (same shape) But shift the graph upwards along the y-axis such that the y=intercept of y = keˣ + 4 is 4½, instead of ½ .
.
.
.
k = ½
bcos x = 0, y = ½
(ii)
eˣ + x/k - 2/k = 0
re-arranging, ...
keˣ = (2 - x)
this is equivalent to simultaneous soln of the following eqns:
y = keˣ . . . . . . . . . [1]
y = 2 - x . . . . . . . . . [2]
if we draw the graph of y = 2-x on the graph of y = keˣ ,
we shld obtain 1 intersection pt.
hence, one solution to eˣ + x/k - 2/k = 0
(iii)
to sketch y = keˣ + 4,
draw the same graph of y = keˣ (same shape) But shift the graph upwards along the y-axis such that the y=intercept of y = keˣ + 4 is 4½, instead of ½ .
.
.
.
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Something like this would be good.
Date Posted:
4 years ago
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