Question

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

Answer 1

There are several ways to find the vertex of a parabola:
1. Complete the square
2. Use `frac{-b}{2a}`
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+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 = `x^2` + 4x + 4
y + 9 = `(x +2)^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}{2a}` from your standard form equation as given. That is, a = 1, b = 4 and c = -5.
x = `frac{-4}{(2*1)}` = `frac{-4}{2}` = -2.

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

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

Method 3. This quadratic is factorable: `x^2+4x-5=(x+5)(x-1)`
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: