f(x)=2x²-8x+6
Find the AXIS OF SYMMETRY and VERTEX. PLACE THE VERTEX AS THE MIDDLE ROW.
Number up and down, then complete the table.
GRAPH!
f(x)=2x²-8x+6 -------> is a vertical parabola open upward
The vertex represent a minimum
Convert the quadratic equation into vertex form
y=a(x-h)^2+k
where
(h,k) is the vertex
Complete the square
f(x)=2x²-8x+6
Factor 2
f(x)=2(x^2-4x)+6
Complete the square
f(x)=2(x^2-4x+4-4)+6
f(x)=2(x^2-4x+4)+6-8
f(x)=2(x^2-4x+4)-2
REwrite as perfect squares
f(x)=2(x-2)^2-2
therefore
The vertex is the point (2,-2)
In a vertical parabola, the axis of symmetry is equal to the x coordinate of the vertex
so
x=2
using a graphing tool
see the attached figure
please wait a minute
we have
f(x)=2x²-8x+6
where
a=2
b=-8
c=6
the formula to calculate the axis of symmetry is
x=-b/2a
substitute the given values
x=-(-8)/(2*2)
x=8/4
x=2