# 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