What is the vertex of the parabola whose equation is #y=x^2+6x+2#?

2 Answers
John C.
May 25, 2017

refer below

Explanation:

The vertex of the parabola can be the maxima or the minima.
This means the gradients at these turning points are zero.

To find the gradient, differentiate the equation to find out the value.

#y=x^2+6x+2#

#y'=2x+6#

When #y'=0#,

#2x+6 = 0#
#x=-3#

Substituting #x=-3# into #y(x)#,

#y=(-3)^2+6(-3)+2#

#y=9-18+2#
#=-7#

Coordinates of the vertex is #(-3,-7)#.

I believe the question does not specify to identify if this is a minima or maxima. If required, just find the second derivative, #y''#. If it is positive, it is a minima and if it is a negative, it is a maxima.

Jim G.
May 25, 2017

#(-3,-7)#

Explanation:

#"for a parabola in the form " y=ax^2+bx+c#

#"the x-coordinate of the vertex is"#

#x_(color(red)"vertex")=-b/(2a)#

#"here " a=1,b=6" and "c=2#

#rArrx_(color(red)"vertex")=-6/2=-3#

#"substitute this value into y and evaluate"#

#rArry_(color(red)"vertex")=(-3)^2+6(-3)+2=-7#

#rArrcolor(magenta)"vertex "=(-3,-7)#
graph{x^2+6x+2 [-20, 20, -10, 10]}

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