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

Question

f(x) = x√(36 − x^2) + c; P(5, 10)

Answer 1

5 * sqrt( 36 - 5^2) + c = 10

5 * sqrt( 36 - 25) + c = 10

5*sqrt( 36 - 25) = 10 - c

5 * sqrt(11) = 10 - c

25*11 = (10 - c)^2

275 = 100 - 20c + c^2

0 = c^2 - 20c - 175

c = [20 +or- sqrt( 400 + 4* 175)]/2

c = [ 20 +or- sqrt( 1100)/2

= [ 20 +or- 10*sqrt(11)]/2

= 10 +or- 5 * sqrt(11)

Answer 2

This problem can by solved by plugging in the coordinates they gave you for x and y and solving for c. Remember f(x) is the same thing as y. In the point P, 5 is our x value and 10 is our y-value:

10 = 5√(36 − 5^2) + c

10 = 5√(36 − 25) + c

10=5√(9) + c

10 = 5 * 3 + c

10 = 15 + c

-15 -15

-5 = c

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