I'm really stuck how to Graph x+y=5 on the coordinate system below showing the graph of y=#4/x#. At what points do the graphs intersect?

1 Answer
Feb 9, 2018

#(1,4)" and "(4,1)#

Explanation:

#"to graph "x+y=5" find the intercepts"#

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

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

#x=0to 0+y=5rArry=5larrcolor(red)"y-intercept"#

#y=0tox+0=5rArrx=5larrcolor(red)"x-intercept"#

#"plot "(0,5)" and "(5,0)#

#"and draw a straight line through them"#

#"to graph "y=4/x#

#"x cannot equal zero as this would make y undefined"#

#x=0" is a vertical asymptote"#

#y!=0" so there are no x-intercepts"#

#"choose some values for x and calculate y"#

#x=1toy=4/1=4rArr(1,4)" is a point"#

#x=2toy=4/2=2rArr(2,2)" is a point"#

#x=4toy=4/4=1rArr(4,1)" is a point"#

#"draw a smooth curve through these points for graph"#

#color(blue)"find the intersection using algebraic method"#

#"substitute "y=4/x" into "x+y=5#

#rArrx+4/x=5#

#"multiply through by x and equate to zero"#

#rArrx^2-5x+4=0larrcolor(blue)"factorise"#

#rArr(x-1)(x-4)=0#

#x-1=0rArrx=1#

#x-4=0rArrx=4#

#"find corresponding y coordinates"#

#x=1toy=4" and "x=4toy=1#

#rArr"points of intersection are "(1,4)" and "(4,1)#
graph{(y+x-5)(y-4/x)((x-1)^2+(y-4)^2-0.04)((x-4)^2+(y-1)^2-0.04)=0 [-10, 10, -5, 5]}