Functions argument help
[From: ] [author: ] [Date: 11-04-23] [Hit: ]
putting the values of x, or the argument, = 6, and 4; you will obtain two equations in a,b,c;36a+6b+c=0 and 16a+4b+c= -8 .......
But if 0 is a zero then f(0)=c=0 which is incorrect.
From the answers y=2x^2-16x+24 and this has zeros when x=2 and x=6.
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Argument is nothing but the value of x, ie f(4)=a4^2 +b4 +c=-8.
So, now in this problem, putting the values of x, or the argument, = 6, and 4; you will obtain two equations in a,b,c;
36a+6b+c=0 and 16a+4b+c= -8 .............A
Now, for the third condition, it is given that the value of this function is minimum at x=4. Since this is an open upward parabola, indirectly the vertex of this parabola is given, because an open upward parabola has minimum value at the vertex.
Now, for finding the vertex in terms of a and b, use the method of completing the square;
f(x) = a{x+b/2a}^2 + (4ac-b^2)/4a.
So, f(x) will be minimum at x= -b/2a = 4,
ie b= -8a.
Also, (4ac-b^2)/4a = -8.
or c = 16a-8
Putting these values of b and c in any one of the equations in A,
you get a=2, then using this value of a in the above two equations,
you get b= -16 and c=24.
Hope this helps :-)
