Question

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

Answer 1

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

Explanation:

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

Answer 2

See below:

Explanation:

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