In your problem you should mention equation of quadratic.
f(x) = aX^{2} + bx +c
Y intercept is always = c , because f(0) = a(0)^{2} + b(0) + c = c
Every Quadratic of aX^{2} + bX + c, with roots (X intercept) of m.n equals
f(x) = k ( X m)( Xn)
So : quadratic of the problem:
k( X +1) ( X3) = k ( X^{2} 2X 3 )
To find the value of K:
2.5 = k ( 0 ^{2} 2( 0)  3) / coordinate of Y intercept ( 0, 2.5)
k = 2.5/ 3 =  5/6
f(x) = 5/6 X^{2} + 5/3 X + 5/2
In the vertex form:
K ( X  b/ 2a) ^{2} ±f ( b/2a)/ also X _{Vertex =}( 3 1)/2 =1
5/6 ( X  1 ) ^{2 }+ f(1) =5/6 ( X1)^{2} + 10/3
If you are not familiar with functional notation :
f(1) = value of Y at X=1
12/11/2013

Parviz F.
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