Question 1090823
{{{f(b)=-b^2-2b+3}}}


choose values for {{{b}}}, calculate {{{f(b)}}}(same as {{{f(x)}}} or {{{y}}}), and make table

{{{b}}}|{{{f(b)}}}

{{{-5}}}|{{{-12}}} ->{{{f(-5)= -(-5)^2-2(-5)+3=-25+10+3=-12}}}

{{{-2}}}|{{{3}}} ->{{{f(-2)= -(-2)^2-2(-2)+3=-4+4+3=3}}}

{{{-1}}}|{{{4}}} ->{{{f(-1)= -(-1)^2-2(-1)+3=-1+2+3=4}}}

{{{0}}}|{{{3}}} ->{{{f(0)= -(0)^2-2(0)+3=3}}}

{{{1}}}|{{{0}}} ->{{{f(1)= -(1)^2-2(1)+3=-1-2+3=0}}}

{{{2}}}|{{{-5}}} ->{{{f(2)= -(2)^2-2(2)+3=-4-4+3=-5}}}

{{{3}}}|{{{-12}}} ->{{{f(3)= -(3)^2-2(3)+3=-9-6+3=-15+3=-12}}}

plot these points and draw a line:


{{{drawing( 600, 600, -10, 10, -20, 10,
circle(-2,3,.13),circle(-1,4,.13),circle(0,3,.13),circle(-5,-12,.13),
circle(1,0,.13),circle(2,-5,.13),circle(3,-12,.13),
locate(1,0,p(1,0)),locate(2,-5,p(2,-5)),locate(3,-12,p(3,-12)),
locate(-2,3,p(-2,3)),locate(-1,4,p(-1,4)),locate(0,3,p(0,3)),locate(-5,-12,p(-5,-12)),
 graph( 600, 600, -10, 10, -20, 10, -x^2-2x+3)) }}}