Question 949449
{{{8x^2-11x-11=0}}}...here you have a parabola (function is of degree {{{2}}})

to use the graphical method, choose several values for variable {{{x}}}, calculate {{{y}}}, you will get the coordinates of the points that lie on this parabola

{{{x}}}|{{{y}}}

{{{-2}}}|{{{43}}} ->{{{8x^2-11x-11=y}}}->{{{8(-2)^2-11(-2)-11=y}}}->{{{32+22-11=y}}}->{{{43=y}}}

{{{-1}}}|{{{8}}} ->{{{8(-1)^2-11(-1)-11=y}}}->{{{8+11-11=y}}}->{{{8=y}}}

{{{0}}}|{{{-11}}} ->{{{8(0)^2-11(0)-11=y}}}->{{{0+0-11=y}}}->{{{-11=y}}}

{{{1}}}|{{{-14}}} ->{{{8(1)^2-11(1)-11=y}}}->{{{8-11-11=y}}}->{{{-14=y}}}

{{{2}}}|{{{-1}}} ->{{{8(2)^2-11(2)-11=y}}}->{{{32-22-11=y}}}->{{{-1=y}}}

{{{3}}}|{{{-1}}} ->{{{8(3)^2-11(3)-11=y}}}->{{{72-33-11=y}}}->{{{28=y}}}

now, plot these points and draw a line through


{{{drawing( 600, 600, -15, 15, -15, 50,
circle(-2,43,.14),circle(-1,8,.14),circle(0,-11,.14),circle(1,-14,.14),circle(2,-1,.14),circle(3,28,.14),
locate(-2,43,p(-2,43)),locate(-1,8,p(-1,8)),
locate(0,-11,p(0,-11)),locate(1,-14,p(1,-14)),
locate(2,-1,p(2,-1)),locate(3,28,p(3,28)),
 graph( 600, 600, -15, 15, -15, 50, 8x^2-11x-11)) }}}