How do you graph #y=(x+5)/(2x-4)# using asymptotes, intercepts, end behavior?
2 Answers
Asymptotes:
Explanation:
Cross multiplying and reorganizing,
(x-2)(y-1/2)=7/2# that represents a rectangular hyperbola (RH) having
asymptotes given by
The center of the RH is (2, 1/2).
y-intercept ( x = 0 ): 5/2
x-intercept ( y = 0 ):
asymptotes
graph{y(2x-4)-x-5=0 [-10, 10, -5, 5]}
See explanation.
Explanation:
A vertical asymptote occurs at
V.A. when
#2x-4=0#
#<=>2x=4#
#<=>x=2#
So the equation for the V.A. is
A horizontal asymptote occurs when the degree of the numerator is less than (or equal to) the degree of the denominator. ("Degree" means the highest power of
When the degrees are the same (like in this case), the horizontal asymptote is found at
(In the case that the denominator has a higher degree, the asymptote is always
The
#0=(x+5)/(2x-4)#
#0=x+5# [multiply both sides by (2x-4) ]
#x="-5"#
So our
Similarly, the
#y=((0)+5)/(2(0)-4)#
#y=5/"-4"=-5/4#
So our
With all this information, we can now draw our hyperbola:
graph{(y-(x+5)/(2x-4))(y-(x-2.0001)/(2x-4))=0 [-8.835, 11.165, -3.91, 6.09]}