# What is the vertex of #f(x)=x^2+4x-5#?

##### 1 Answer

There are several ways to find the vertex of a parabola:

1. Complete the square

2. Use

3. Find the average of the two x-intercepts, and substitute that x-value in to find the y-value. (f(x))

4. Use a graphing calculator and the "analyze" feature

- Here is how I would complete the square:

Replace f(x) with "y" and make sure that the lead coefficient = 1

# y = x^2+4x - 5#

#y + 5 = x^2 + 4x#

y + 5 +**???_ =**_???__#x^2# + 4x +

(take half of the coefficient of the x-term, then square it)

so y + 5 + 4 =

y + 9 =

Then set y + 9 = 0 and x + 2 = 0 to get y = -9 and x = -2 which are the coordinates of the vertex (-2, -9).

Method 2 involves calculating

x =

Now that you have the x-value for the vertex, substitute it into the function and find f(-2) =

Confirmed, (-2,-9) is the vertex.

Method 3. This quadratic is factorable:

So the zeros are x = -5 and 1. Average the zeros:

Method 4 involves using technology and not showing any work to your instructor. Is this really how you want to do it? Here is what my calculator would show: