Question 32040
{{{y = x^2-4x-5}}}

a) Convert to{{{y = a(x-h)^2+k}}}form.
{{{y = x^2-4x-5}}} Add 5 to both sides.
{{{y+5 = x^2-4x}}} Complete the square in the x-terms by adding the square of half the x-coefficient {{{(4/2)^2 = 4}}}to both sides.
{{{y+5+4 = x^2-4x+4}}} Factor the right side. Simplify the left side.
{{{y+9 = (x-2)^2}}} Finally, subtract 9 from both sides.
{{{y = (x-2)^2-9}}}
Compare this with {{{y = a(x-h)^2+k}}}
a = 1, h = 2, and k = -9

b) The line of symmetry (LOS) is the vertical line through the center of the vertex. The vertex is located at (h, k), so in this problem, the vertex is located at (2, -9).
The LOS is therefore the vertical line x = 2

c)& d) The graph of{{{y = x^2-4x-5}}}(in red).
The graph of {{{y = x^2}}}(in green).
{{{graph(300,200,-6,6,-10,5,x^2-4x-5,x^2)}}}
As you can see, the two parabolas have the same shape but the red one (your equation) is translated to the right by h units (h=2) and down k units (k = -9). So by first graphing the the parabola{{{y = x^2}}} whose vertex lies at the origin, then translating (sliding) it so that the vertex is at the location defined by (h, k) in your equation, you really don't need to plot points.