Question

How do you find the vertex of a quadratic equation?

Answer 1

Use the formula `-b/(2a)` for the x coordinate and then plug it in to find the y.

Explanation:

A quadratic equation is written as `ax^2+bx+c` in its standard form. And the vertex can be found by using the formula `-b/(2a)`.

For example, let's suppose our problem is to find out vertex (x,y) of the quadratic equation `x^2+2x-3` .

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 `-b/(2a)`. For this example, you'll get `-2/(2*1)` which can be simplified to -1.

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) `(-1)^2+2(-1)-3=y`.

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.

Answer 2

`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 `f(x)=0`:

`=> ax^2+bx+c = 0`

With roots, `alpha` and `beta`.

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`-coordinate of the vertex is:

`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 `x`:

`f'(x) = 2ax+b`

At a critical point, the first derivative , `f'(x)` vanishes, which requires that:

`f'(x) = 0`

`:. 2ax+b =0`
`:. x = -b/(2a) \ \ \ \` QED