How do you write the vertex form equation of the parabola #y=-x^2+2x+3#?

1 Answer
Jun 4, 2017

Complete the square.

Explanation:

For #y=-x^2 + 2x +3#, take out the factor of #-1# to make the #x^2# coefficient #+1#.

#:. y=-(x^2-2x-3)#

Now, complete the square. Divide the #x# coefficient by #2# and square it, adding it and subtracting it.

#:. y=-(x^2-2x+1-1-3)#
#:. y=-((x-1)^2-4)#
#:. y=-(x-1)^2 + 4#

This is now in turning-point form; simply read off the coordinates of your vertex, which is at #(1,-4)#. Remember that this is a local maximum. Of course, the original equation factorises quite easily, so for sketching purposes, make #y=-(x-3)(x+1)# to get your #x#-intercepts; the #y#-intercept is, of course, just the constant term in the original equation.