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

Yes. It would not be a function if there were $\left(a , b\right) , \left(a , c\right) , b \setminus \ne c$

Explanation:

Three points define a parabola $y = a {x}^{2} + b x + c$

First point:
$4 = a {\left(- 2\right)}^{2} - 2 b + c \setminus R i g h t a r r o w c \left(a , b\right) = 4 - 4 a + 2 b$

Second point:
$5 = a {\left(1\right)}^{2} + b + c$

$\setminus R i g h t a r r o w 5 = a + b + 4 - 4 a + 2 b$

$\setminus R i g h t a r r o w 1 = - 3 a + 3 b$

$\setminus R i g h t a r r o w \frac{1}{3} + a = b \left(a\right) \setminus R i g h t a r r o w c \left(a\right) = 4 - 4 a + 2 \left(\frac{1}{3} + a\right)$

$\setminus R i g h t a r r o w c \left(a\right) = \frac{14}{3} - 2 a$

Third point:
$- 2 = a {\left(5\right)}^{2} + 5 b + c$

$\setminus R i g h t a r r o w - 2 = 25 a + 5 \left(\frac{1}{3} + a\right) + \frac{14}{3} - 2 a$

$\setminus R i g h t a r r o w - 6 = 75 a + 5 \left(1 + 3 a\right) + 14 - 6 a$

$\setminus R i g h t a r r o w - 20 = 69 a + 5 + 15 a$

$\setminus R i g h t a r r o w - \frac{25}{84} = a$

$\setminus R i g h t a r r o w b = \frac{1}{3} - \frac{25}{84} = \frac{3}{84}$

$\setminus R i g h t a r r o w c = \frac{14}{3} - 2 \left(- \frac{25}{84}\right) = \frac{28 \cdot 14 + 50}{84} = \frac{442}{84}$

$y = f \left(x\right) = \frac{- 25 {x}^{2} + 3 x + 442}{84}$

Exercise: Determine $a , b , r$ in the circle ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

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