Question 760887
f(x)=x^2-4x-5 i need help finding y intercept, x intercept, vertex, and axis of symmetry
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y=x^2-4x-5
To find y-intercept, set x=0, then solve for y:
y=0-0-5
y-intercept=-5
..
To find x-intercepts, set y=0 then solve for x:
0=x^2-4x-5
factor:
(x-5)(x+1)=0
x=5
and 
x=-1
x-intercepts are 5 and -1
..
To find the vertex:
Complete the square to write equation in standard form:
y=x^2-4x-5
y=(x^2-4x+4)-4-5
y=(x-2)^2-9
This is an equation of a parabola that opens upward.
Its standard (vertex) form:y=(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex
vertex: (2,-9)
axis of symmetry: x=2
see graph below as a visual check:
{{{ graph( 300, 300, -10, 10, -10, 10, x^2-4x-5) }}}