Question

How do you find the vertex of `y=x^2+10x+21`?

Answer 1

`"vertex" = (-5, -4)`

Explanation:

`x=-b/(2a)`

`x=-10/(2(1))`

`x=-5`

Sub `-5` into the equation

`y=(-5)^2 + 10(-5) + 21`

`y = -4`

The formula `-b/(2a)` is used to find the axis of symmetry which is
always the `x` value of the vertex. Once you find the `x` value of the vertex, you simply substitute that value into the quadratic equation and find the `y` value, which in this case, is the vertex.

Answer 2

(-5,-4)

Explanation:

You have to use the quadratic formula `x=(-b+-sqrt(b^2-4ac))/2a`
which becomes
`x=-b/(2a)+-(sqrt(b^2-4ac)/(2a))`
We know that `-b/(2a)` is constant and that the other part is plussing and minusing from it

So it is the vertex and as `a=1 b=10 c=21` i.e. just the coefficents of all the terms in sequence.

The vertex must be `-10/(2*1)` so the x co-ordinate of the vertex is `-5`

Plug in `f(-5)` and you get the y co-ordinate
`f(-5)=(-5)^2+10(-5)+21` becomes `f(-5)=25-50+21`
so `f(-5)=-4`

so the co-ordiantes of the vertex are (-5,-4)