Posted by Maria1 on Thursday, April 26, 2012 at 11:49am.
f(x)=1x^2.
Looking for the vertex and an equation of the line of symmetry.
f(x)=x^2+5x+36 Looking for the x, and Y coordinates.
I seem to be entering the answers backwards, cannot understand.
Please help.

Algebra  Reiny, Thursday, April 26, 2012 at 12:26pm
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
(x9)(x+4) = 0
x = 9 or x = 4
for yintercept let x = 0
then y = f(0) = 36
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