What is the vertex, axis of symmetry, the maximum or minimum value, and the range of parabola #y=-x^2+2x-5#?

All you Need can you see in the following formula
#y=-(x^2-2x+1-1)-5=-(x-1)^2-4#

To find the answers in a more straightforward manner, let's first convert from standard form to vertex form. To do that, we're changing from

#y=Ax^2+Bx+C#

into

#y=(x-h)^2+k#

To do that, we complete the square:

#y=-x^2+2x-5#

#y=-(x^2-2x)-5#

#y=-(x^2-2x+1)-5+1#

#y=-(x-1)^2-4#

Vertex

In this form, the vertex (or, in other words, the "pointy bit") is given to us in the #(h,k)# term. Here we have #(1,-4)#

Axis of Symmetry

The axis of symmetry splits the parabola in 2 equal parts. It runs through the vertex straight up and down (for a parabola that either opens up or opens down). And so we can write an equation for the line that is vertical and runs through the vertex. In this question, it's #x=1#.

Maximum/minimum value

Here we're dealing with looking at the vertex and seeing if it is as high as the parabola goes (i.e. maximum and is when there is a negative sign sitting in front of the #(x-h)^2# term) or as low as it goes (i.e. minimum and is when there is a positive sign sitting there).

In our case, there's a negative sign and so this will be a maximum. The #y# coordinate of the vertex shows the value.

#y=-4#

Range

Since we know the maximum value is #y=-4#, all #y# values less than and including #-4# will be part of the parabola graph.

#y<=-4#

To see all of this in graph form, here's the graph:

graph{-x^2+2x-5[-10,10,-10,0]}

For more on parabolas, you may find this helpful:

http://jwilson.coe.uga.edu/emt725/class/sarfaty/emt669/instructionalunit/parabolas/parabolas.html