So we have #y = -x^2 + 2x +4#, and usually the sorts of 'important' points that are standard for including on sketches of quadratics are axis intercepts and the turning points.
To find the #x#-intercept, simply let #y=0#, then:
#-x^2 + 2x +4 = 0#
Then we complete the square (this will also help with finding the turning point).
#x^2 - 2x + 1# is the perfect square, then we subtract one again to maintain the equality:
#-(x^2 - 2x + 1) + 1 +4 = 0#
#:. -(x-1)^2 + 5 = 0#
This is the 'turning-point' form of the quadratic, so you can read your stationary point right off: #(1,5)# (alternatively you could differentiate and solve #y' = 0#).
Now just transpose the equation:
#(x-1)^2 = 5#
#:. x- 1 = +- sqrt5#
#:. x = 1+-sqrt5#
The #y#-intercept is easy, When #x=0#, #y = 4#.
And there you have it!
