標題:
唔識做功課38
發問:
the graph of y=-x^2+8x-7 intersects the x-axis at the point A and B. M(p , q) is a point on the graph such that q>0. a) Find the coordinates of A and B. b) Express the area of triangle AMB in terms of q. c) Find the maximum area of triangle AMB and the corresponding coordinates of M.
最佳解答:
It may be a little bit messy, sorry about that. a) 0=-x^2+8x-7 0=x^2-8x+7 x=(-(-8)+sqrt((-8)^2-4*1*7))/(2*1) or (-(-8)-sqrt((-8)^2-4*1*7))/(2*1) x=(8+sqrt(64-28))/2 or (8-sqrt(64-28))/2 x=(8+sqrt(36))/2 or (8-sqrt(36))/2 x=(8+6)/2 or (8-6)/2 (these steps are just using the quadratic equation) x=7 or 1 the coordinates of A and B are (7,0) or (1,0) b) the area of triangle AMB =AB*height*0.5 =AB*q*0.5 (Because q>0, M lies on Quadrant I or II, So that the height is the value of q (q-0) ) =(7-1)*q*0.5 =3q c)To find the max area of the triangle, we must the max height as the base can't be changed y=-x^2+8x-7 y=-(x^2-8x+7) y=-(x^2-8x+16-16+7) y=-((x-4)^2-9) y=-(x-4)^2+9 so that the highest point is (4,9) (I suppose you know what's going on) the max area =3q =3*9 =27 sq.unit If you still have questions, please feel free to ask.
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