Posted by Maria1 on Thursday, April 26, 2012 at 11:49am.
shortcut to find vertex of
y = ax^2 + bx + c
the x of the vertex is -b/(2a)
in your case, the x of the vertex = 0/-2 = 0
sub back in
f(x) = 1
vertex is (0,1)
axis of symmetry is x = 0
2nd:
you must mean the x and y -intercepts
let y or f(x) = 0
x^2 - 5x - 36 = 0
(x-9)(x+4) = 0
x = 9 or x = -4
for y-intercept let x = 0
then y = f(0) = 36
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