SOLUTION: Can someone help me with this problem? I'm stuck!!! I'm in a class that is over my head. Help!!!!!!!!!!!!!!!!!!!! For the function y = x2 - 6x + 8, perform the following tas

Algebra ->  Linear-equations -> SOLUTION: Can someone help me with this problem? I'm stuck!!! I'm in a class that is over my head. Help!!!!!!!!!!!!!!!!!!!! For the function y = x2 - 6x + 8, perform the following tas      Log On


   

Question 79162: Can someone help me with this problem? I'm stuck!!! I'm in a class that is over my head. Help!!!!!!!!!!!!!!!!!!!!
For the function y = x2 - 6x + 8, perform the following tasks:
a) Put the function in the form y = a(x - h)2 + k.
Answer:
Show work in this space.


b) What is the equation for the line of symmetry for the graph of this function?
Answer:



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.
Show graph here.

Explanation of graphing.

d) In your own words, describe how this graph compares to the graph of y = x2?
Answer:

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
a) Put the functiony+=+x%5E2-6x%2B8 into the form y+=+a%28x-h%29%5E2%2Bk
To do this you will start with the given function and you will "complete the square".
y+=+x%5E2-6x%2B8 Now subtract the constant (8) from both sides.
y-8+=+x%5E2-6x Now complete the square in the x-terms by adding the square of half the x-coefficient (that's %28-6%2F2%29%5E2+=+9) to both sides.
y%2B1+=+x%5E2-6x%2B9 Factor the right side.
y%2B1+=+%28x-3%29%5E2 Now subtract 1 from both sides.
y+=+%28x-3%29%5E2-1 Compare this with:
y+=+a%28x-h%29%5E2%2Bk
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.
%28x-3%29%5E2-1+=+0 Add 1 to both sides.
%28x-3%29%5E2+=+1 Take the square root of both sides.
sqrt%28x-3%29%5E2+=+sqrt%281%29
x-3+=+1 + or - Add 3 to both sides.
x+=+3%2B1 and x+=+3-1
x+=+4 and x+=+2 These are the x-intercepts.
y-intercept: Set x = 0 and solve for y.
y+=+%280-3%29%5E2+-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%5E2 (in green) so you can compare the two.
graph%28300%2C200%2C-5%2C6%2C-4%2C10%2C%28x-3%29%5E2-1%2Cx%5E2%29
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.