IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: xxemoxx on December 10, 2009, 10:21:12 am
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If curve y=p-(x-q)^2 crosses x-axis at (-1,0) and (5,0). Find maximum values of y.
another question
What values of m makes the following expression in perfect square.
(2m-1)x^2 - 4(m-2)x + (m-1)
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For the first one,
first thing u need to do is find the values of p and q, although p is not really necessary, but we shall find them anyways
substitute in the values for x and y into
y = p - (x - q)²
At (-1,0)
0 = p - ( -1 - q )²
0 = p - (q² + 2q + 1)
p = q² + 2q + 1
At (5,0)
0 = p - ( 5 - q )²
0 = p - (25 - 10q +q²)
p = q² - 10q + 25
Equate both the p's
q² - 10q + 25 = q² + 2q + 1
Giving,
- 12q = -24
q = 2
q² + 2q + 1 = p
(2)²+2(2) + 1 = p
4 + 4 + 1 = p
p = 9
now we know that
y = 9 - (x - 2)²
to find the maximum values, you need to differentiate
dy/dx = [USING CHAIN RULE]
======> -2(x-2)(1)
=> -2(x-2)
at maximum values, dy/dx = 0
therefore
-2(x-2) = 0
x- 2 = 0
x = 2
y = 9 - (2 - 2)²
y = 9
therefore the coordinates of maximum are (2,9)
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And the second one,
for any equation to be a "perfect square" there needs to be only one root
therefore,
b² = 4ac
our equation, in its simplified form
(2m-1)x² + (-4m +8) x + (m-1)
b = (-4m+8)
a = (2m-1)
c = (m-1)
b² = 4ac
(-4m+8)² = 4(2m-1)(m-1)
16m² - 64m + 64 = 4(2m²-3m + 1)
4m² - 16m + 16 = 2m² - 3m + 1
2m² - 13m + 15 = 0
(2m - 3)(m - 5) = 0
Therefore,
the values at which m is a perfect square is when m = 1½ and when m = 5