Posted by **Maria1** on Thursday, April 26, 2012 at 11:49am.

f(x)=1-x^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

(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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