Answered

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find real numbers a, b, and c so that the graph of the function y=ax^2+c contains the points (-1,6), (2,7), and (0,1)

Sagot :

ollieboyne
This gives you three simultaneous equations:

6 = a + c
7 = 4a + c
1 = c
 
c = 1

If c =1,

6 = a + 1
a = 5

This doesn't work in the second equation, so the quadratic that goes through these points is not in the form y = ax^2 + bx + c
Was there supposed to be a b in the equation?




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