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

##### 2 Answers

Jul 22, 2018

#### 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]}

Jul 22, 2018

#### Explanation:

The given function:

Above equation is in the standard form of vertical parabola:

Which has **vertex** at