How do you graph #y=x+4# by plotting points?

2 Answers
Dec 16, 2017

See a solution process below:

Explanation:

For a linear function you only need to plot two points. First, solve for two points which solve the equation and plot these points:

First Point: For #x = 0#

#y = 0 + 4#

#y = 4# or #(0, 4)#

Second Point: For #x = 2#

#y = 2 + 4#

#y = 6# or #(2, 6)#

We can next plot the two points on the coordinate plane:

graph{(x^2+(y-4)^2-0.1)((x-2)^2+(y-6)^2-0.1)=0 [-20, 20, -10, 10]}

Now, we can draw a straight line through the two points to graph the line:

graph{(y-x-4)(x^2+(y-4)^2-0.1)((x-2)^2+(y-6)^2-0.1)=0 [-20, 20, -10, 10]}

Dec 16, 2017

#"see explanation"#

Explanation:

#"one way is to find the intercepts, that is where the"#
#"graph crosses the x and y axes"#

#• " let x = 0, in the equation for y-intercept"#

#• " let y = 0, in the equation for x-intercept"#

#x=0toy=0+4rArry=4larrcolor(red)"y-intercept"#

#y=0to0=x+4rArrx=-4larrcolor(red)"x-intercept"#

#"plot the points "(0,4)" and "(-4,0)#

#"and draw a straight line through them"#
graph{(y-x-4)((x-0)^2+(y-4)^2-0.04)((x+4)^2+(y-0)^2-0.04)=0 [-10, 10, -5, 5]}