4 years ago

Find the value of c such that the point P(a, b) lies on the graph of the function f.

1 answer:

4 0

$f(x)=x\sqrt{16-x^2}+c;\ P(3;\ 6)\to x=3\ and\ y=6\\\\subtitute\\\\6=3\sqrt{16-3^2}+c\\\\3\sqrt{16-9}+c=6\\\\3\sqrt7+c=6\ \ \ \ |subtract\ 3\sqrt7\ from\ both\ sides\\\\\boxed{c=6-3\sqrt7}$

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