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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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:

`0=a(8)^2+b(8)`

`0=64a+8b`

`64a=-8b`

`a=-b/8` (2)

The equation of the tangent is:

`y=16x+b` (3)

Substitute 2 for x in (3):

`y=32+b`

Subsstitute y=32+b in (1):

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

`4a+b=32` (4)

Substitute (2) in (4):

**b=16**

Substitue 16 for b in (2):

`a=-16/8=-2`

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

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