Match the function with its graph.1) y= tan x2) y= cot x3) y= -tan x4) y= -cot x

Answer:Option a. 1C, 2A, 3B, 4D.Step-by-step explanation:1) We know that tan(x)=sin(x)/cos(x). If x=0, sin(x)=0 and cos(x)=1 then tan(x)=0. For that reason, we know that the graph passes through the point (0,0). If x=45, then sin(45)= [tex]\frac{\sqrt{2}}{2}[/tex] and cos(45)=[tex]\frac{\sqrt{2}}{2}[/tex]. Thus tan(45)=1. The only graph that passes through the point (0,0) and is possitive when x=45 is the graph C.2) We know that cot(x)=cos(x)/sin(x). If x=0, sin(x)=0 and cos(x)=1 then tan(x)=+∞. For that reason, we know that the graph has an asymptote in y=0, in other words, it never crosses the y-axis. If x=45, then sin(45)= [tex]\frac{\sqrt{2}}{2}[/tex] and cos(45)=[tex]\frac{\sqrt{2}}{2}[/tex]. Thus cot(45)=1. The only graph that has an asymptote in y=0 and is possitive when x=45 is the graph A. 3) We know that -tan(x)=-sin(x)/cos(x). If x=0, sin(x)=0 and cos(x)=1 then -tan(x)=0. For that reason, we know that the graph passes through the point (0,0). If x=45, then sin(45)= [tex]\frac{\sqrt{2}}{2}[/tex] and cos(45)=[tex]\frac{\sqrt{2}}{2}[/tex]. Thus -tan(45)=-1. The only graph that passes through the point (0,0) and is negative when x=45 is the graph B.) We know that -cot(x)=-cos(x)/sin(x). If x=0, sin(x)=0 and cos(x)=1 then tan(x)=-∞. For that reason, we know that the graph has an asymptote in y=0, in other words, it never crosses the y-axis. If x=45, then sin(45)= [tex]\frac{\sqrt{2}}{2}[/tex] and cos(45)=[tex]\frac{\sqrt{2}}{2}[/tex]. Thus -cot(45)=-1. The only graph that has an asymptote in y=0 and is negative when x=45 is the graph D.