Question

How do I find the vertex of `f(x)=x^2+6x+5`?

Answer 1

`"vertex "=(-3,-4)`

Explanation:

`"given the parabola in "color(blue)"standard form"`

`•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0`

`"then the x-coordinate of the vertex is"`

`•color(white)(x)x_(color(red)"vertex")=-b/(2a)`

`f(x)=x^2+6x+5" is in standard form"`

`"with "a=1,b=6" and "c=5`

`x_("vertex")=-6/2=-3`

`"substitute this value into the equation for y"`

`y_("vertex")=(-3)^2+6(-3)+5=-4`

`rArrcolor(magenta)"vertex "=(-3,-4)`
graph{x^2+6x+5 [-10, 10, -5, 5]}

Answer 2

`(-3, -4)`

Explanation:

The given function:

`f(x)=x^2+6x+5`

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

`y=(x+3)^2-4`

`(x+3)^2=(y+4)`

Above equation is in the standard form of vertical parabola:

`(x-x_1)^2=4a(y-y_1)`

Which has vertex at

`(x_1, y_1)\equiv(-3, -4)`