Find a so that the point (1,2) is on the graph of f (x)=ax^2+5

a = -5
Heyy, so with this question you would have to substiute 1 for x because normally the coordinates have x first then y. E.g. (x,y)

soo, f(1) = a(1)^2 +5
then, f(1) = a + 5
then equate to zero. E.g. a+5=0
then move over 5 to other side E.g. a=-5
soo.. yah.. pretty sure haha :)