Question 1152348
first recall equation in the slope-intercept form:

{{{y=mx+b}}} where {{{m}}} is a slope and {{{b}}} is y-intercept

Graph the following equations using the slope-intercept method. 
Draw the y-intercept in yellow. 
Draw at least one additional point in red using the slope:


{{{y =-2x+highlight_green(2)}}}

=> y-intercept is at ({{{0}}}{{{highlight_green(2)}}}) (I am using green where should be yellow because yellow here does not work)


one additional point: 

if {{{y=0}}}=>{{{0 = -2x +2}}}=>{{{2x=2}}}=>{{{x=highlight(1)}}}
=> x-intercept is at ({{{highlight(1)}}}{{{0}}})


{{{drawing( 600, 600, -10, 10,-10, 10,
circle(0,2,.12), locate(0,2,p(0,highlight_green(2))),
circle(1,0,.12), locate(1,-0.5,p(highlight(1),0)),
 graph( 600, 600, -10, 10, -10, 10, -2x +2)) }}} 


{{{y = (3/4)x - 1}}}

=>y-intercept is at ({{{0}}}{{{highlight_green(-1)}}})

one additional point: 

if {{{y=0}}}=>{{{0= (3/4)x - 1}}}=>{{{1= (3/4)x }}}=>{{{x= 1/(3/4)=4/3}}}

=> x-intercept is at ({{{highlight(4/3)}}}{{{0}}})

{{{drawing( 600, 600, -10, 10,-10, 10,
circle(0,-1,.12), locate(-0.9,-1,p(0,highlight_green(-1))),
circle(4/3,0,.12), locate(4/3,1.2,p(highlight(4/3),0)),
 graph( 600, 600, -10, 10, -10, 10, (3/4)x - 1)) }}} 



a line that goes through the points 

({{{2}}}, {{{-3}}}) and ({{{2}}},{{{ 4}}})

 If you plotted these points, what would this graph look like? 

{{{drawing( 600, 600, -10, 10,-10, 10,
circle(2,-3,.12), locate(2,-3,p(2,-3)),
circle(2,4,.12), locate(2,4,p(2,4)),
 graph( 600, 600, -10, 10, -10, 10, 0)) }}} 

What should this equation be?

as you can see both points have same {{{x}}} coordinate, and both lie on vertical line {{{x=2}}}

you can also use a slope to determine what should this equation be:

{{{m=(y[2]-y[1])/(x[2]-x[1])=(4-(-3))/(2-2)=7/0}}}=> no slope, means we have vertical line


{{{drawing( 600, 600, -10, 10,-10, 10,
circle(2,-3,.12), locate(2,-3,p(2,-3)),
circle(2,4,.12), locate(2,4,p(2,4)),green(line(2,10,2,-10)),
 graph( 600, 600, -10, 10, -10, 10, 0)) }}}