Question

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

Answer 1

`(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]}