parabola y=4-(x)^2 cuts x axis at R and S... P (x,y) lies on parabola in 1st quadrant, and Q lies on parabols such that PQ is paralel to the x-axis...
i already worked out R is (-2,0) and S is (2,0)
now i need to show area of trapeziumb PQRS is given by:
A = (2-x)(4-x^2)
aaargh!!??? wat!? Area of trapezium = 1/2(a + b)h... but not much help?
then i need to find value of x which gives the maximum value of A... but i think i can work that out if i get this.
pls help, thanx heaps :)
2007-12-02
20:10:29
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3 answers
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asked by
Anonymous