What is the vertex of #f(x)=x^2+4x-5#?
There are several ways to find the vertex of a parabola:
1. Complete the square
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
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: