How do you graph #y=x+4# by plotting points?
2 Answers
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
Second Point: For
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]}
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]}