Method 1: convert the equation into vertex form
Note: vertex form is y=color(green)m(x-color(red)a)^2+color(blue)b for a parabola with vertex at (color(red)a,color(blue)b)
y=2(x-1)^2-4xcolor(white)("xxxxxxxx")...as given
expanding
y=2(x^2-2x+1)-4x
y=2(x^2-2x+1-2x)
y=2(x^2-4x+1)
completing the square
y=2(x^2-4x+4)-6
we added 3 to the previous 1 but this is multiplied by 2 so we need to subtract 2xx3=6 to keep this equivalent.
y=color(green)2(x-color(red)2)^2+color(blue)(""(-6))
which is the vertex form with vertex at (color(red)2,color(blue)(-6))
Method 2: Note that the slope (derivative) of the parabola at the vertex is zero
y=2(x-1)^2-4x
expanding:
y=2x^2-8x+2
at the vertex
y'=4x-8 = 0
color(white)("XXX")rArr color(red)(x=2) at the vertex
Substituting 2 for x back in the original equation gives
color(blue)y=2(2-1)^2-4 * 2= 2-8color(blue)(=-6)
Again, giving the vertex at
color(white)("XXX")(color(red)2,color(blue)(-6))
Method 3: Use a graphing calculator/software package
![enter image source here]()
