turning point of a parabola (method 2)

To find the turning point of a parabola use the formula 

`x=(-b)/(2a)`

Then put the value of `x` back into the quadratic and you will be able to find the coordinate of `y` on the turning point.

Example 1

Find the turning point of

`x^2-6x+8=y`

use the formula `x=(-b)/(2a)`

Remember that the general formula for a quadratic is

`ax^2+bx+c=y`

`x=(-(-6))/(2times1)=(6)/(2)=3`

`x=3`

Now put the value of `x=3` back into the quadratic equation.

`x^2-6x+8=y`

`3^2-6`x`3+8=y`

`9-18+8=y`

`y=-1`

Example 2

Find the turning point of 

`y=(x+3)^2+5`

Use the formula `x=(-b)/(2a)`

Remembering that the general formula for a quadratic is 

`ax^2+bx+c=y`

So we first must turn `y=(x+3)^2+5` into the above format

`y=(x+3)(x+3)+5`

`y=x^2+3x+3x+9+5`

`y=x^2+6x+14`

Now using `x=(-b)/(2a)`

`x=(-6)/(2times1)`

`x=-3`

Now put the value of `x=3` back into the quadratic equation

`y=(-3)^2+6(-3)+14`

`y=9-18+14`

`y=+5`

The turning point occurs at

`x=-3` and `y=5`

or `(-3,5)`