Is {(-2,4), (1,5),(5,-2)} a function?

1 Answer

Yes. It would not be a function if there were #(a,b), (a,c), b \ne c#

Explanation:

Three points define a parabola #y = ax^2 + bx + c#

First point:
#4 = a(-2)^2 -2 b + c \Rightarrow c(a,b) = 4 - 4a + 2b#

Second point:
#5 = a(1)^2 + b + c #

#\Rightarrow 5 = a + b + 4 - 4a + 2b#

#\Rightarrow 1 = -3a + 3b#

#\Rightarrow 1/3 + a = b(a) \Rightarrow c(a) = 4 - 4a + 2(1/3 + a)#

#\Rightarrow c(a) = 14/3 - 2a#

Third point:
#-2 = a(5)^2 + 5b + c#

#\Rightarrow -2 = 25a + 5(1/3 + a) + 14/3 - 2a#

#\Rightarrow -6 = 75a + 5(1 + 3a) + 14 - 6a#

#\Rightarrow -20 = 69a + 5 + 15a#

#\Rightarrow -25/84 = a#

#\Rightarrow b = 1/3 - 25/84 = 3/84#

#\Rightarrow c = 14/3 - 2(-25/84) = (28 * 14 + 50)/84 = 442/84#

#y = f(x) = (-25x^2 + 3x + 442)/84#

Exercise: Determine #a,b,r# in the circle #(x - a)^2 + (y - b)^2 = r^2#

The three points are in the upper semicircle or in the downer one?