1. The graph of the quadratic function y=f(x)=ax^2+bx+c. The vertex of the graph is at(3,0) and the y-intercept is 18.
(a) Show that b^2=72a.
Thx~
Maths about graph
2009-07-26 6:27 am
回答 (2)
2009-07-26 6:42 am
✔ 最佳答案
y=f(x)=ax^2+bx+c, vertex of the graph is at(3,0) and the y-intercept is 18.This means that f(0)=18=>c=18
Rewrite f(x) as ax^2+bx+c=a(x^2+b/ax+c/a)=a[(x-b/2a)^2+c/a-b^2/4a^2]=a[(x-b/2a)^2+(b^2-4ac)/4a^2]
So (b^2-4ac)/4a=0 (since the vertex is (3,0) )
b^2=4ac=72a
2009-07-26 8:46 am
a(x^2+b/ax+c/a)=a[(x-b/2a)^2+c/a-b^2/4a^2] E個 step 唔明, May you explain to me, thx so much~
2009-07-28 19:04:26 補充：
Could anyone explain me?Thx~ I don't why the ans. is that.
2009-07-28 19:04:26 補充：
Could anyone explain me?Thx~ I don't why the ans. is that.
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