How do you graph #y = sqrt (3x) + 4# by plotting point?

1 Answer
Oct 7, 2017

#y=sqrt(3x)+4#
graph{sqrt(3x)+4 [-10.38, 9.62, -5, 5]}

#y=sqrt3x+4#
graph{sqrt3x+4 [-10.38, 9.62, -5, 5]}

Explanation:

1) If given equation is #y=sqrt(3x)+4#
#x=0,y=4#
#x=1,y=sqrt3+4#
#x=2,y=sqrt6+4#
#x=3,y=7#
Four points are #(0,4),(1,5.732),(2,6.449),(3,7)#

2) If given equation is #y=sqrt3*x+4#
Giving values for x, find corresponding value of y from the above equation and plot in a graph.
#x=0,y=4#
#x=1,y=sqrt3+4#
#x=2,y=2sqrt3+4#
#x=-1,y=-sqrt3+4#
Four points are #(0,4),(1,5.732),(2,7.464),(-1,2.268)#