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

Sep 4, 2014

There are several ways to find the vertex of a parabola:
1. Complete the square
2. Use $\frac{- b}{2 a}$
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

1. Here is how I would complete the square:
Replace f(x) with "y" and make sure that the lead coefficient = 1
$y = {x}^{2} + 4 x - 5$
$y + 5 = {x}^{2} + 4 x$
y + 5 + ???_ = ${x}^{2}$ + 4x + _???__

(take half of the coefficient of the x-term, then square it)
so y + 5 + 4 = ${x}^{2}$ + 4x + 4
y + 9 = ${\left(x + 2\right)}^{2}$ (the right side is a perfect square trinomial!)
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 $\frac{- b}{2 a}$ from your standard form equation as given. That is, a = 1, b = 4 and c = -5.
x = $\frac{- 4}{\left(2 \cdot 1\right)}$ = $\frac{- 4}{2}$ = -2.

Now that you have the x-value for the vertex, substitute it into the function and find f(-2) = ${\left(- 2\right)}^{2} + 4 \left(- 2\right) - 5$ = 4 + (-8) - 5 = -9.

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

Method 3. This quadratic is factorable: ${x}^{2} + 4 x - 5 = \left(x + 5\right) \left(x - 1\right)$
So the zeros are x = -5 and 1. Average the zeros: $\frac{- 5 + 1}{2}$ = $\frac{- 4}{2}$ = -2. Repeat finding f(-2) to get -9.

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: