请教数学paper 1

lonely_world

Given y=ax&#178;+bx+c,a.b &c are postive constants. Show that the least value of y is -&#8710;/4a where &#8710;=b&#178;-4ac, & find the corresponding value of x. Sketch the graph for (1)&#8710;>0,(2)&#8710;<0

请问这题怎样做?

Log

在maths t paper 1 讨论区

lonely_world

原帖由 Log 于 9-9-2009 02:50 PM 发表
在maths t paper 1 讨论区

知道了..
可是还有问题..

Log

what problem?

idontwant2b

y = ax^2 + bx + c
   = a(x^2 + b/a x) +c
   = a[x^2 + b/a x + (1/2)^2(b/a)^2 - (b/2a)^2]+c
   = a[x + (b/2a)]^2 - a(b/2a)^2 + c
   = a[x+ (b/2a)]^2 - a (b/2a)^2 + c
   = a[x+ (b/2a)]^2 + (c - b^2 /4a)
   = a[x+ (b/2a)]^2 + (4ac-b^2)/4a
   = a[x+ (b/2a)]^2 + - (b^2 - 4ac)/4a
   = a[x+ (b/2a)]^2 + - D/4a

minimum of y = -D/4a
corresponding x  = -a/2b

for D>0, 2 real roots, sketch any parabola graph on the straight line with two points touching the straight line

for D<0, 2 complex roots, sketch any parabola graph on the straight line with no intersection with the straight line,

这样对吗？

Log

yes,but你是不是写错了呢？corresponding x= -b/2a
not -a/2a

Log

其实不是any parabola.因为given a,b and c are +ve integers
so a>0, parabola = U-shaped curve