What is the vertex of # y=x^2/7-7x+1 #?

1 Answer
Jun 11, 2018

#(24.5,-84.75)#

Explanation:

#y= =>a=1/7,b=-7,c=1#
for co-ordinate of vertex #(h,k)#
#h=-b/(2a)=7/(2.(1/7))=49/2#
put #x=49/2# to find #y# and corresponding point #k#
#k=-84.75#
co-ordinate is #(24.5,-84.75)#

best method : by calculus
vertex is the lowermost(or uppermost) point #i.e# minimum or maximum of the function
we have
#y=x^2/7-7x+1#
#=>(dy)/(dx)=2x/7-7#
at minimum or maximum slope of curve is 0 or #(dy)/(dx)=0#
#=>2x/7-7=0=>x=49/2#

check if this point is of maximum or minimum by second derivative test(thisstep is not necessarily needed)
if second derivative is -ve it corresponds to point of maximum
if second derivative is +ve it corresponds to point of minimum

#(d^2y)/(dx^2)=2/7=+ve=>x=49/2# corresponds to point of minimum
now put #x=49/2# to find #y#
and you will find coordinates as
#(24.5,-84.75)#
and it's evident from the graph

graph{x^2/7-7x+1 [-10, 10, -5, 5]}