Question

What is the vertex of `y=x^2+12x+26`?

Answer 1

The vertex is at `(-6, -10)`

Explanation:

You can find the vertex (turning point) by first finding the line that is the axis of symmetry.

`x = (-b)/(2a) = (-12)/(2(1)) = -6" "larr` This is the `x`-value of the vertex.

Now find `y`.

`y = x^2 +12x+26`

`y = (-6)^2 +12(-6) +26`
`y= 36-72+26`
`y=-10" "larr` This is the `y`-value of the vertex.

The vertex is at `(-6, -10)`

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You can also find the vertex by completing the square to get the equation in vertex form: `y = a(x+b)^2 +c`

`y = x^2 +12x+26`
`y= x^2 +12x color(red)(+6^2) color(red)(-6^2) +26" " color(red)((b/2)^2 =(12/2)^2)`

`y = (x+6)^2 -10`

Vertex is at `(-b, c)" "rarr (-6, -10)`