Given the quadratic function f(x)=x^2-2x-8 find the following information algebraically (show all work), then sketch the graph.

X intercept is where the function hits the x axis or where y=0
y intercept is where the function hits the y axis or where x=0
axis of symmetry is x=c where c is a constant, it is also the x value of the vertex
the vertex is the lowest or highest point on the parabola

to find the x intercept, solve for f(x)=0
0=x^2-2x-8
factor
0=(x-4)(x+2)
set to zero
x=-2 and 4
so x intercepts at (-2,0) and (4,0)

y intercepts
set x=0
f(0)=0^2-2(0)-8
f(0)=-8
y intercept at (0,-8)

to find the axis of symmetry in form
f(x)=ax^2+bx+c
the value of the constant is -b/(2a)
so
f(x)=1x^2-2x-8
axis of symmetry is -(-2)/(2*1)=2/2=1
so it is x=1

vertex, subsitut the x value to find the y value
x=1
f(1)=1^2-2(1)-8
f(1)=1-2-8
f(1)=-9
so vertex is (1,-9)

x intercepts are (-2,0) and (4,0)
y intercept at (0,-8)
axis of symmetry is x=1
vertex is (1,-9)

Where is your question?