Question 46964
a) Put the function in the form y = a(x - h)2 + k.
y = x^2 - 6x + 8
v(-b/2a,f(x))
v(3,-1) so h = 3 and k = -1
a = 1 while b = -6 and c = 8
{{{y = a(x - h)^2 + k}}}
{{{y = (x - 3)^2 - 1}}}
b) What is the line of symmetry?
Since the parabola is vertical, the line of symmetry is equal to the x-term in the vertex. {{{x = 3}}}
c) Graph the function using the equation in part a. Explain why it is not necessary to plot points to graph when using y = a (x - h) 2 + k.
{{{y = (x - 3)^2 - 1}}} in terms: {{{y = a(x - h)^2 + k}}}
vertex(h,k)
(distance from vertex to focus or from vertex to directrix = {{{1/4a}}} = {{{1/4}}}
Latus Rectum (distance touching the parabola going through the focus) = |1/a| = 1
{{{graph(600,600,-10,10,-10,10,(x - 3)^2 - 1)}}}
d) In your own words, describe how this graph compares to the graph of y = x2?
They are both parabolas. They both have the same value for 'a'. {{{y = x^2}}} has a different vertex. The Latus Rectum and distances are the same.