`f(x)= 2x^2-4x -5`

First, we will rewrite using the vertex form `f(x)= (x-h)^2 + k`

`==> f(x)= 2x^2 -4x -5+2 -2`

`==> f(x)= (2x^2 -4x +2) -5 -2`

`==> f(x)= 2(x^2-2x +1) -7`

`` `==> f(x)= 2(x-1)^2 -7`

`==> h= 1`

**Then, the axis of symmetry is x = 1**

`==> k = -7`

Now we will find the vertex.

`V = (v_x, v_y)`

`==> V = (h, k) = (1, -7)`

Using the standard form, the vertex is:

`v_x = -b/(2a) = 4/4 = 1`

`v_y= -(b^2-4ac)/(4a)= -(16+40)/8 = -56/8 = -7`

`==> V = (1,-7)`

Now we will find the y-intercept:

We know that the y-intercept is the point where the function meets the y-axis. Then, the value of x is 0.

`==> x = 0`

`==> f(0)= -5`

**Then, the y-intercept is the point (0, -5).**

** **

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.