Find the vertex of the parabola formed by the following quadratic function: y = 3x^2 – 6x + 2

Vertex form is y=a(x-h)^2+k, so we can rearrange to that form...

y=3x^2-6x+2  subtract 2 from both sides

y-2=3x^2-6x  divide both sides by 3

(y-2)/3=x^2-2x, halve the linear coefficient, square it, add it to both sides...in this case:  (-2/2)^2=1 so

(y-2)/3+1=x^2-2x+1  now the right side is a perfect square

(y-2+3)/3=(x-1)^2

(y+1)/3=(x-1)^2  multiply both sides by 3

y+1=3(x-1)^2  subtract 1 from both sides

y=3(x-1)^2-1  so the vertex is:

(1, -1)

...

Now if you'd like you can commit to memory the vertex point for any parabola so you don't have to do the calculations like what we did above.  The vertex of any quadratic (parabola), ax^2+bx+c is:

x= -b/(2a),   y= (4ac-b^2)/(4a)

Then you will always be able to do a quick calculation of the vertex :)

Distance=slopeslope is -2/4which equals-1/2the distance between these two points is -1/2