#color(blue)("Determine the x-intercepts")#
Note that 23 is a prime number and as the coefficient of #x# IS NOT #+1-23=-22# the roots will be fractional. Thus use the formula
Given that: #y=ax^2+bx+c" where "x=(-b+-sqrt(b^2-4ac))/(2a)#
In this case: #a=+1"; "b=-1"; "c=-23#
#x=(+1+-sqrt((-1)^2-4(1)(-23)))/(2(1))#
#x=(1+-sqrt(93))/2#
93 is not prime so we can 'hunt' for squared values as factors of it. The whole number factors turn out to be 3 and 31. Both of which are prime so we are stuck with #sqrt(93)# as the exact value.
Thus #x=1/2+-sqrt(93)/2" "larr# exact values
#" "x~~-4.321825..." and "x~~5.321825...#
#" "x~~-4.32" and "x~~5.32# to 2 decimal places
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the vertex")#
#x_("vertex")# will be midpoint between the x-intercepts
#x_("vertex") ->(-4.32+5.32)/2=0.5#
As a check this is a 'sort of cheat method'
From #y=ax^2+bx+c:" "x_("vertex")=(-1/2)xxb#
#" "-> (-1/2)xx(-1)=+0.5#
#y_("vertex")=x^2-x-23" "=" "(0.5)^2-0.5-23#
#" "=-23.25 =-93/4#
Vertex#->(x,y)=(1/2,-93/4) ->(0.5,-23.25)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the y-intercept")#
Set #x=0#
#y=x^2-x-23-=>y=0^2+0-23#
#y_("intercept")=-23#
