Question 1048919
{{{f(x)=x^2 -x - 48}}}

first write it in vertex form {{{f(x)=a(x-h)^2 +k}}} where {{{h}}} and {{{k}}} are {{{x}}} and {{{y}}} coordinates of the vertex

{{{f(x)=(x^2 -x) - 48}}}...complete square

{{{f(x)=(x^2 -x+b^2)-b^2 - 48}}}.....since coefficient {{{a=1}}} and {{{2ab=-1}}}, find {{{b}}}
{{{2*1b=-1}}}->{{{2b=-1}}}->{{{b=-1/2}}}

so, we have
{{{f(x)=(x^2 -x-(-1/2)^2)-(-1/2)^2 - 48}}}

{{{f(x)=(x-1/2)^2-(1/4) - 48}}}

{{{f(x)=(x-1/2)^2-1/4 - 192/4}}}

{{{f(x)=(x-1/2)^2 - 193/4}}}


so, {{{h=1/2}}} and {{{k=-193/4}}}

or {{{h=0.5}}} and {{{k=-48.25}}}


set {{{f(x)=0}}} to find {{{x}}}-intercepts

{{{0=(x-1/2)^2 - 193/4}}}

{{{193/4=(x-1/2)^2 }}}

{{{sqrt(193/4)=x-1/2 }}}

{{{1/2+sqrt(193)/2=x }}}

{{{x}}}-intercepts are: 

{{{x=1/2+sqrt(193)/2}}} or {{{x=1/2-sqrt(193)/2 }}}

{{{x=7.5}}} or {{{x=-6.5)}}}



set {{{x=0}}} to find {{{y}}}-intercepts

{{{f(x)=(0-1/2)^2 - 193/4}}}

{{{f(x)=(-1/2)^2 - 193/4}}}

{{{f(x)=1/4 - 193/4}}}

{{{f(x)= - 192/4}}}

{{{f(x)= - 48}}} which is same if you use {{{f(x)=x^2 -x - 48}}}->{{{f(x)=0^2 -0 - 48=-48}}}

{{{x=7.5}}} or {{{x=-6.5)}}}
{{{drawing( 600, 600, -25, 25, -55, 50,
circle(1/2,-193/4,.42), locate(1/2,-193/4,V(1/2,-193/4)),
 circle(7.5,0,.42),circle(-6.5,0,.42),
 graph( 600, 600, -25, 25, -55, 50, x^2 -x - 48)) }}}