How do you find the vertex of #f(x)= x^2+10x-8#?

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Answer 1 · 2015-06-07T13:36:00

To find the vertex you can complete the square:

#f(x) = x^2+10x-8#

#= (x^2+10x+25)-25-8#

#= (x+5)^2-33#

The vertex is where the term #(x+5)^2# has its minimum possible value, #0#, which is when #x=-5# and #f(x) = -33#

That is, the vertex is at #(-5, -33)#

In general:

#ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))#