Question

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.

Answer 1

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