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

May 25, 2018

$y = - {\left(x - 1\right)}^{2} - 4$

#### Explanation:

All you Need can you see in the following formula
$y = - \left({x}^{2} - 2 x + 1 - 1\right) - 5 = - {\left(x - 1\right)}^{2} - 4$

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

$y = A {x}^{2} + B x + C$

into

$y = {\left(x - h\right)}^{2} + k$

To do that, we complete the square:

$y = - {x}^{2} + 2 x - 5$

$y = - \left({x}^{2} - 2 x\right) - 5$

$y = - \left({x}^{2} - 2 x + 1\right) - 5 + 1$

$y = - {\left(x - 1\right)}^{2} - 4$

Vertex

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

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 ${\left(x - h\right)}^{2}$ 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.

$y \le - 4$

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