PLzz help I will give u more points don't spam or llf
2 answers:
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Answer:
Check the attached graph
Step-by-step explanation:
Given equation is .
Now we need to graph the given equation ..
Clearly it is not linear as equation is not of the form . So we need to find some points on given equation
. by plugging some random values of x like -5,-3,1,... etc.
for x=-5, we get:
.
Hence first point is at (-5,-2). Similarly we can find more points and graph those points.
x can't be 0 as division by 0 is not defined so it will have vertical asymptote at x=0.
Since degree of numerator is less than degree of denominator so y=0 will be horizontal asymptote.
Now join those points by a curved line to get final graph.
