Question 486981
    {{{f(x)=x^2-2x-24}}}

a - Determine the vertex?

Vertex Form of Equation  
The vertex form of a parabola's equation is generally expressed as :

{{{y= a(x-h)^2+k}}}

    (h,k) is the vertex

{{{f(x)=x^2-2x-24}}}

{{{x^2-2x+1 -25}}}

{{{(x^2-2x+1) -25}}}

{{{(x-1)^2-25}}}

so,  (h,k) is  (1,-25) 

b - Does the graph open up or down?

If a> 0, the parabola opens upwards
        if a< 0, it opens downwards.

{{{a=1}}}.. {{{a> 0}}}, the parabola opens upwards

c - What is the equation of the axis of symmetry?

 The axis of symmetry is the line {{{x = -b/2a}}}
{{{a=1}}} and  {{{b=-2}}}

{{{x = -(-2)/2*1}}}

{{{x = 2/2}}}

{{{x = 1}}}


d - Find any x-intercepts?

{{{f(x)=x^2-2x-24}}}....set {{{y=0}}}


{{{x^2-2x-24=0}}}...replace {{{-2x}}} with {{{4x-6x}}}

{{{x^2+4x-6x-24=0}}}...group


{{{(x^2+4x)-(6x+24)=0}}}


{{{x(x+4)-6(x+4)=0}}}

{{{(x-6)(x+4)=0}}}

if  {{{x-6=0}}}.....->...{{{x=6}}}.......one x-intercept is at (6,0)



if  {{{x+4=0}}}.....->...{{{x=-4}}}.......other one x-intercept is at (-4,0)



e - Find the y-intercepts?

{{{f(x)=x^2-2x-24}}}....set {{{x=0}}}

{{{f(x)=0^2-2*0-24}}}

{{{f(x)=-24}}}......
so, y-intercept is at (0,-24)


f - Sketch the graph.

{{{ graph( 500, 500, -25,25, -30, 20, x^2-2x-24) }}}


g - State the domain and range in interval notation? 

*[invoke Plot_A_Graph_With_Domain "x^2-2x-24", -30, 30, -30, 30]