Dde asked in 科學及數學數學 · 1 decade ago

amatha

Consider a quadratic curve C:y=-x+(h-9)x+(10h+10),where h is greater than or equal to 0

(a)Find the vertex of C

(b)If the point lies on C,express p in terms of h

(c)If the curve does not intersect with the horizontal line y=100,find th range of possible values of h

1 Answer

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  • 1 decade ago
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    (a) By completing the square

    y = -x^2 + (h - 9)x + (10h + 10)

    = - [ x^2 + (h - 9)x + (h - 9)/2 - (h - 9)/2 ] + (10h + 10)

    = - [ x + (h-9)/2 ]^2 + (h - 9)/2 + (10h + 10)

    = - [ x+ (h-9)/2 ]^2 + (21h + 11)/2

    So, the vertex of C is ( -(h - 9)/2 , (21h + 11)/2 )

    (b) unclear definition

    (c) put y = 100

    100 = -x^2 + (h - 9)x + (10h + 10)

    x^2 - (h - 9)x + (90 - 10h) = 0

    delta = (h - 9)^2 - 4 (90 - 10h) < 0

    h^2 - 18h + 81 - 360 + 40h < 0

    h^2 + 22h - 279 < 0

    (h - 9)(h + 31) < 0

    -31 < h < 9

    since h >= 0,

    the possible range of h is 0 <= h < 9

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