How do you find the vertex of #f(x) = ( 2p ) x ^ { 2} - ( 12p ) x + ( 3n )#?

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Answer 1 · 2017-09-14T02:36:22

#f_"Vertex" = (3, -18p+3n)#

Here we have a quadratic in standard form: #ax^2+bx+c#

Expressed as : #(2p)x^2-(12p)x+(3n)#

#:. a=2p, b=-12p, c=3n# Where {p,n} are assumed to be real constants.

We know that the vertex of the parabola defined by the quadratic will lie on the axis of symmetry, where #x=(-b)/(2a)#

I.e. where #x = (12p)/(2xx2p)#

#= (12cancelp)/(2xx2cancelp) = 3#

Hence, the #x-#component of the vertex will be #3#

To find the #y-#component we must find the value of the quadratic at #x=3#

#f(3) = 2pxx(3)^2 - 12pxx3 +3n#

#= 18p - 36p +3n#

#= -18p+3n# which is the #y-#component of the vertex.

Hence, the vertex of #f(x)# is the point #(3, -18p+3n)#