Question 935219
{{{f(x)=1.5x+3}}} and {{{f(x)=-x^2 +4}}}


1.

{{{f(x)=1.5x+3}}}

make a table:

{{{x}}}|{{{f(x)=1.5x+3}}}...since linear, we need just two points

{{{-2}}}|{{{0}}}
{{{0}}}|{{{3}}}


{{{drawing( 600, 600, -10, 10, -10, 10,

circle(-2,0,.14),locate(-2,0,p(-2,0)),
circle(0,3,.14),locate(0,3,p(0,3)),

 graph( 600, 600, -10, 10, -10, 10,1.5x+3)) }}}


2.

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

{{{-2}}}|{{{0}}}
{{{-1}}}|{{{3}}}
{{{0}}}|{{{4}}}
{{{1}}}|{{{3}}}
{{{2}}}|{{{0}}}
{{{-3}}}|{{{-5}}}
{{{3}}}|{{{-5}}}


plot points and draw a line through:


{{{drawing( 600, 600, -10, 10, -10, 10,

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