Find numbers a, b, and c so that the graph of `f(x) = a(x)^(2)+b(x)+c` has x-intercepts at (0,0) and (8,0), and a tangent with slope 16 where x=2.



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Posted on (Answer #1)

Substitute x=0 and y=0 in`f(x)=y=ax^2+bx+c`   (1)

` `c=0

Substitute x=8 and y=0 and c=0:




`a=-b/8`    (2)

The equation of the tangent is:

`y=16x+b` (3)

Substitute 2 for x in (3):


Subsstitute y=32+b in (1):

`32+b=4a+2b` or

`4a+b=32`  (4)

Substitute (2) in (4):


Substitue 16 for b in (2):


Thus the function is:`f(x)=-2x^2+16x`

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