Question

What is the vertex of `y = x^2+10x+21`?

Answer 1

In the standard form `y=ax^2+bx+c` the `x`-coordinate of the vertex is `-b/(2a)`

In this situation `a=1`, `b=10` and `c=21`, so the `x`-coordinate of the vertex is:
`-b/(2a)=-10/(2xx1) = -5`

Then we simply substitute `x=-5` into the original equation to find the `y`-coordinate of the vertex.

`y=(-5)^2+10(-5)+21=-4`

So the coordinates of the vertex are:
`(-5, -4)`