How do you find the vertex of # y=x^2+10x+21#?

2 Answers
Adrian C. · Stefan V.
Mar 26, 2018

#"vertex" = (-5, -4)#

Explanation:

#x=-b/(2a)#

#x=-10/(2(1))#

#x=-5#

Sub #-5# into the equation

#y=(-5)^2 + 10(-5) + 21#

#y = -4#

The formula #-b/(2a)# is used to find the axis of symmetry which is
always the #x# value of the vertex. Once you find the #x# value of the vertex, you simply substitute that value into the quadratic equation and find the #y# value, which in this case, is the vertex.

Rohan N.
Mar 26, 2018

(-5,-4)

Explanation:

You have to use the quadratic formula #x=(-b+-sqrt(b^2-4ac))/2a#
which becomes
#x=-b/(2a)+-(sqrt(b^2-4ac)/(2a))#
We know that #-b/(2a)# is constant and that the other part is plussing and minusing from it

So it is the vertex and as #a=1 b=10 c=21# i.e. just the coefficents of all the terms in sequence.

The vertex must be #-10/(2*1)# so the x co-ordinate of the vertex is #-5#

Plug in #f(-5)# and you get the y co-ordinate
#f(-5)=(-5)^2+10(-5)+21# becomes #f(-5)=25-50+21#
so #f(-5)=-4#

so the co-ordiantes of the vertex are (-5,-4)

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