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

1 Answer
Ayush S.
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]}

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