How do you find the vertex of a quadratic equation?
2 Answers
Use the formula
Explanation:
A quadratic equation is written as
For example, let's suppose our problem is to find out vertex (x,y) of the quadratic equation
1) Assess your a, b and c values. In this example, a=1, b=2 and c=-3
2) Plug in your values into the formula
3) You just found the x coordinate of your vertex! Now plug in -1 for x in the equation to find out the y-coordinate.
4)
5) After simplifying the above equation you get : 1-2-3 which is equal to -4.
6) Your final answer is (-1 ,-4)!
Hope that helped.
# ax^2+bx+c = 0 # has a vertex at#(-(b)/(2a), -(b^2 - 4ac)/(4a) )#
Explanation:
Consider a general quadratic expression:
# f(x) = ax^2+bx+c = 0 #
and its associated equation
# => ax^2+bx+c = 0 #
With roots,
We know (By symmetry - See below for proof) that the vertex (either maximum or minimum) is the mid-point of the two root, the
# x_1 = (alpha+beta)/2 #
However, recall the well studied properties:
# {: ("sum of roots", = alpha+beta, = -b/a), ("product of roots", = alpha beta, = c/a) :} #
Thus:
# x_1 = -(b)/(2a) #
Giving us:
# f(x_1) = a(-(b)/(2a))^2+b(-(b)/(2a))+c #
# \ \ \ \ \ \ \ \ = (b^2)/(4a) - b^2/(2a)+c #
# \ \ \ \ \ \ \ \ = (4ac - b^2)/(4a) #
# \ \ \ \ \ \ \ \ = -(b^2 - 4ac)/(4a) #
Thus:
# ax^2+bx+c = 0 # has a vertex at#(-(b)/(2a), -(b^2 - 4ac)/(4a) )#
Proof of midpoint:
If we have
# f(x) = ax^2+bx+c = 0 #
Then, differentiating wrt
# f'(x) = 2ax+b #
At a critical point, the first derivative ,
# f'(x) = 0 #
# :. 2ax+b =0 #
# :. x = -b/(2a) \ \ \ \ # QED