Question

Graph the function. Identify the vertex and axis of symmetry.
f(x)=-2x^2+2x-1

Answer 1

Part A) The vertex is the point
`(0.5,-0.5)`

Part B) The axis of symmetry is
`x=0.5`

Part C) The graph in the attached figure

Explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

`y=a(x-h)^(2)+k`

where

(h,k) is the vertex of the parabola

and the axis of symmetry is equal to the x-coordinate of the vertex

so

`x=h` -----> equation of the axis of symmetry

In this problem we have

`f(x)=-2x^(2)+2x-1`

Convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`f(x)+1=-2x^(2)+2x`

Factor the leading coefficient

`f(x)+1=-2(x^(2)-x)`

Complete the square. Remember to balance the equation by adding the same constants to each side

`f(x)+1-(1/2)=-2(x^(2)-x+(1/4))`

`f(x)+(1/2)=-2(x^(2)-x+(1/4))`

Rewrite as perfect squares

`f(x)+(1/2)=-2(x-(1/2))^(2)`

`f(x)=-2(x-(1/2))^(2)-(1/2)` -----> equation in vertex form

The vertex of the parabola is the point
`(0.5,-0.5)`

Is a vertical parabola open downward

The axis of symmetry is equal to

`x=0.5`

see the attached figure to better understand the problem

Answer 2

We will be expressing the equation in vertex form:
y = a(x - h)² + k, where the vertex is (h , k)
y = -2(x² - x + 1/2)
y = -2(x² - x + 1/4 - 1/4 + 1/2)
y = -2(x² - 1/2) - 1/2

Vertex = (1/2 , -1/2)
Axis of symmetry: x = 1/2