Question 120485
given:

{{{x = -4}}}, {{{x=-3}}}, {{{x=-2}}}, {{{x=-1}}}, {{{x=0}}}, {{{x=1}}}, {{{x=2}}}

{{{y=x^2 + 2x - 3}}}


to find:


if {{{x = -4}}}, then {{{y=x^2 + 2x - 3}}} will be:


{{{y=x^2 + 2x - 3}}}......substitute {{{x}}} with {{{-4}}}


{{{y=(-4)^2 + 2(-4) - 3}}}......


{{{y= 16 - 8 - 3}}}......


{{{y= 16 - 11 }}}......


{{{y= 5}}}......



if {{{x = -3}}} then {{{y=x^2 + 2x - 3}}} will be:



{{{y=x^2 + 2x - 3}}}......substitute {{{x}}} with {{{-3}}}


{{{y=(-3)^2 + 2(-3) - 3}}}......


{{{y= 9 - 6 - 3}}}......


{{{y= 9 - 9 }}}......


{{{y= 0}}}......


if {{{x = -2}}}, then {{{y=x^2 + 2x - 3}}} will be:



{{{y=x^2 + 2x - 3}}}......substitute {{{x}}} with {{{-2}}}


{{{y=(-2)^2 + 2(-2) - 3}}}......


{{{y= 4 - 4 - 3}}}......


{{{y= 0 - 3 }}}......


{{{y= -3}}}......


if {{{x = -1}}}, then {{{y=x^2 + 2x - 3}}} will be:



{{{y=x^2 + 2x - 3}}}......substitute {{{x}}} with {{{-1}}}


{{{y=(-1)^2 + 2(-1) - 3}}}......


{{{y= 1 - 2 - 3}}}......


{{{y= 1 - 5 }}}......


{{{y= -4}}}......



if {{{x = 0}}}, then {{{y=x^2 + 2x - 3}}} will be:


{{{y=x^2 + 2x - 3}}}......substitute {{{x}}} with {{{-0}}}


{{{y=(0)^2 + 2(0) - 3}}}......


{{{y= 0 - 0 - 3}}}......


{{{y= -3}}}......



if {{{x = 1}}}, then {{{y=x^2 + 2x - 3}}} will be:



{{{y=x^2 + 2x - 3}}}......substitute {{{x}}} with {{{1}}}


{{{y=(1)^2 + 2(1) - 3}}}......


{{{y= 1 + 2 - 3}}}......


{{{y= 3 - 3 }}}......


{{{y= 0}}}......



if {{{x = 2}}}, then {{{y=x^2 + 2x - 3}}} will be:


{{{y=x^2 + 2x - 3}}}......substitute {{{x}}} with {{{2}}}


{{{y=(2)^2 + 2(2) - 3}}}......


{{{y= 4 + 4 - 3}}}......


{{{y= 8 - 3 }}}......


{{{y= 5}}}......