The figure shows the graphs of y=-(x+p)^2+q which cuts the y-axis at B(0,4). The equation of the line VA is x=3, where V is the vertex of the graph
(a)Find the values of p and q
(b) Hence find the area of the quadrilateral OAVB.
F.5 Maths More about Graphs of
2014-01-04 1:54 am
回答 (1)
2014-01-04 2:21 am
✔ 最佳答案
For y = - ( x + p)^2 + q, the axis of symmetry is x = - p, which is x = 3 because V is the vertex.So - p = 3, p = - 3.
That is y = - ( x - 3)^2 + q.
Since B(0,4) is on the curve,
4 = - (0 - 3)^2 + q
4 = - 9 + q
q = 13
So y = - (x - 3)^3 + 13.
Co - ordinates of V is ( 3, 13).
Since the figure is not available, assume A to be (3,0).
So area of OAVB = [ OB + VA] x OA/2 = (4 + 13) x 3/2 = 17 x 3/2 = 51/2.
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