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# Complete the square and find the vertex form of each quadratic function, then write the vertex and the axis of symmetry and draw the graph. f(x)=x^2-6x+1

I honestly don't understand how to solve this question and need help!

f(x) = x^2 - 6x + 1

f(x) = (x^2 - 6x ) + 1

f(x) = (x^2 - 6x + 9 ) + 1 - 9

f(x) = (x - 3)2 - 8

f(x) = a(x-h)2 + k

Vertex = (3, -8)

Axis of symmetry => x = 3

You could use -b/(2a), which gives you -(-6)/(2*1) = 3

f(3) = 3^2 - 6(3) + 1 = 9 - 18 + 1 = -8

The graph crosses the x axis in two places:

6 +/- sqrt(36-4)
------------------
2

6 +/- sqrt(32)
---------------
2

6 +/- sqrt(16) sqrt(2)
------------------------
2

6 +/- 4 sqrt(2)
----------------
2

x = 3 +/- 2sqrt(2)

Parabola is open upward, with a vertex at (3, -8), crossing the x axis at (3-2sqrt(2), 0), and (3+2sqrt(2), 0).

f ( x ) =  X2  - 6X  + 1

= X2  - 2 ( 6x/ 2)  + 9 = - 1 + 9

( X - 3 )2  = 8                                         Roots are:

( X- 3)2 = 8  ,   X = 3 ±2√2

Vertex : ( 3, 8)

General formula :  ( -b/(2a) ,    (b2 - 4ac) / 4a2 )