Question 906490
{{{f(x)=(x^2-9x+1)/8}}}, As meant when given,


{{{f(x)=(1/8)(x^2-9x+1)}}}, a bit neater to handle;
The factor {{{1/8}}} squashes the parabola vertically.


The square term to use for transforming into standard form is {{{(-9/2)^2=81/4}}}.


{{{f(x)=(1/8)(x^2-9x+1+81/4-81/4)}}}


{{{f(x)=(1/8)(x^2-9x+81/4+1-81/4)}}}
{{{f(x)=(1/8)((x^2-9x+81/4)+1-81/4)}}}
{{{f(x)=(1/8)((x-9/2)^2+4/4-81/4)}}}
{{{f(x)=(1/8)(x-9/2)^2+(1/8)(-77/4)}}}
{{{highlight(f(x)=(1/8)(x-9/2)^2-77/32)}}}----Standard Form.


The vertex, a minimum point of the function, is {{{highlight(x=9/2)}}}, {{{highlight(y=-77/32=-2&13/32)}}}



Set f(x)=0 and solve for x to know the x-intercepts.
Set x=0 and solve for y to know the y-intercept.