Question 79162
a) Put the function{{{y = x^2-6x+8}}} into the form {{{y = a(x-h)^2+k}}}
To do this you will start with the given function and you will "complete the square".

{{{y = x^2-6x+8}}} Now subtract the constant (8) from both sides.
{{{y-8 = x^2-6x}}} Now complete the square in the x-terms by adding the square of half the x-coefficient (that's {{{(-6/2)^2 = 9}}}) to both sides.
{{{y+1 = x^2-6x+9}}} Factor the right side.
{{{y+1 = (x-3)^2}}} Now subtract 1 from both sides.
{{{y = (x-3)^2-1}}} Compare this with:
{{{y = a(x-h)^2+k}}}
a = 1
h = 3
k = -1

b) The equation of the line of symmetry is given by:
{{{x = h}}}, so in this case, since h = 3, the equation of the line of symmetry is:
{{{x = 3}}} This is a vertical line passing through the point (3, 0)

c) It is not necessary to plot points to graph the function in this form because:
1) You know the equation of the line of symmetry.
2) You know that the parabola opens upward because a>0 (a is positive).
3) You know (or can find) the location of the vertex of the parabola.  Thi is given by (h, k) or (3, -1)
4) You can find the x- and y-intercepts of the function by:
x-intercepts: Set y = 0 and solve for x.
{{{(x-3)^2-1 = 0}}} Add 1 to both sides.
{{{(x-3)^2 = 1}}} Take the square root of both sides.
{{{sqrt(x-3)^2 = sqrt(1)}}}
{{{x-3 = 1}}} + or - Add 3 to both sides.
{{{x = 3+1}}} and {{{x = 3-1}}}
{{{x = 4}}} and {{{x = 2}}} These are the x-intercepts.
y-intercept: Set x = 0 and solve for y.
{{{y = (0-3)^2 -1}}}
{{{y = 9-1}}}
{{{y = 8}}} This the y-intecept.

Here is the graph (in red): I'll add the graph of the function {{{y = x^2}}} (in green) so you can compare the two.
{{{graph(300,200,-5,6,-4,10,(x-3)^2-1,x^2)}}}

d) I'll leave the explanation to you.  Look at the two graphs and try to see what effects the values of h and k have on the placement of the second (green) graph compared to the first (red) graph.