Question 681992
1.

{{{f(x)= x^2-2x-8}}}...factor completely, replace {{{-2x}}} with {{{2x-4x}}}

{{{f(x)= x^2+2x-4x-8}}}.......group

{{{f(x)= (x^2+2x)-(4x+8)}}}

{{{f(x)= x(x+2)-4(x+2)}}}

{{{f(x)= (x-4)(x+2)}}}....set {{{f(x)=0}}}


{{{0= (x-4)(x+2)}}}

if {{{0= (x-4)}}}..=>...{{{x=4}}}

if {{{0= (x+4)}}}..=>...{{{x=-2}}}


so, the roots are {{{x=4}}} and {{{x=-2}}}

and points are ({{{4}}},{{{0}}}) and  ({{{-2}}},{{{0}}})

let {{{x=1}}}, solve for  {{{f(x)}}} and find third point


{{{f(1)= 1^2-2*1-8}}}

{{{f(1)= 1-2-8}}}

{{{f(1)= -9}}}

so, third point is ({{{1}}},{{{-9}}})

now, graph it


{{{drawing(600,600,-12,12,-12,12,grid(0),circle(4,0,0.2),circle(-2,0,0.2),circle(1,-9,0.2),graph(600,600,-12,12,-12,12,x^2-2x-8))}}}



2.

{{{f(x)=  x^2-2x-15}}}...factor completely, replace {{{-2x}}} with {{{3x-5x}}}

{{{f(x)= x^2+3x-5x-15}}}.......group

{{{f(x)= (x^2+3x)-(5x+15)}}}

{{{f(x)= x(x+3)-5(x+3)}}}

{{{f(x)= (x-5)(x+3)}}}....set {{{f(x)=0}}}


{{{0= (x-5)(x+3)}}}

if {{{0= (x-5)}}}..=>...{{{x=5}}}

if {{{0= (x+3)}}}..=>...{{{x=-3}}}


so, the roots are {{{x=5}}} and {{{x=-3}}}

and points are ({{{5}}},{{{0}}}) and  ({{{-3}}},{{{0}}})

let {{{x=1}}}, solve for  {{{f(x)}}} and find third point


{{{f(1)= 1^2-2*1-15}}}

{{{f(1)= 1-2-15}}}

{{{f(1)= -16}}}

so, third point is ({{{1}}},{{{-16}}})

now, graph it


{{{drawing(600,600,-10,10,-18,10,grid(0),circle(5,0,0.2),circle(-3,0,0.2),circle(1,-16,0.2),graph(600,600,-10,10,-18,10,x^2-2x-15))}}}