Question 1131316
.	Using the following values for 
{{{x}}}: {{{-2}}},{{{ -1}}}, {{{0}}}, {{{1}}}, {{{2 }}}


a.	plot and connect the points for this equation: 

{{{y = x^2 + 1}}} 

{{{x}}}|{{{y}}}
{{{-2}}}|{{{5}}}->{{{y = (-2)^2 + 1=5}}}
{{{-1}}}|{{{2}}}->{{{y = (-1)^2 + 1=2}}}
{{{0}}}|{{{1}}}->{{{y = (0)^2 + 1=1}}}
{{{1}}}|{{{2}}}->{{{y = (1)^2 + 1=2}}}
{{{2}}}|{{{5}}}->{{{y = (2)^2 + 1=5}}}


{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(-2,5,.12),locate(-2,5,p(-2,5)),
circle(-1,2,.12),locate(-1,2,p(-1,2)),
circle(0,1,.12),locate(0,1,p(0,1)),
circle(1,2,.12),locate(1,2,p(1,2)),
circle(2,5,.12),locate(2,5,p(2,5)),

graph( 600, 600, -10, 10, -10, 10, x^2 + 1)) }}}




b.	plot and connect the points for this equation:	{{{y = x + 1 }}}

{{{x}}}|{{{y}}}
{{{-2}}}|{{{-1}}}->{{{y = -2 + 1=-1}}}
{{{-1}}}|{{{0}}}->{{{y = -1 + 1=0}}}
{{{0}}}|{{{1}}}->{{{y = 0 + 1=-1}}}
{{{1}}}|{{{2}}}->{{{y = 1 + 1=2}}}
{{{2}}}|{{{3}}}->{{{y = 2 + 1=3}}}

{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(-2,-1,.12),locate(-2,-1,p(-2,-1)),
circle(-1,0,.12),locate(-1,0,p(-1,0)),
circle(0,1,.12),locate(0,1,p(0,1)),
circle(1,2,.12),locate(1,2,p(1,2)),
circle(2,3,.12),locate(2,3,p(2,3)),

graph( 600, 600, -10, 10, -10, 10, x + 1)) }}}



c.	the shapes are different. What do you think accounts for that?

{{{y = x^2 + 1}}}  is a 2nd degree function, graph is a parabola
{{{y = x + 1}}} is 1st degree function, graph is a line