How do you graph y=4/(3x-6)+5y=43x−6+5 using asymptotes, intercepts, end behavior?

1 Answer
LM
Dec 20, 2016

graph{(4/(3x-6))+5 [-17.58, 22.42, -6.32, 13.68]}

asymptote: x = 2x=2, vertical

yy-intercept is at (0,4.33)(0,4.33)

xx-intercept is at (1.73,0)(1.73,0)

y<5y<5 when x<2x<2
y>5y>5 when x>2x>2

Explanation:

asymptotes:

n/0n0 = undefined

n/0 +5n0+5 =undefined

asymptote: 4/(3x-6) = 4/043x−6=40

3x-6 = 03x−6=0

3x = 63x=6

x = 2x=2

intercepts:

yy:
yy-intercept: x = 0x=0

y = 4/(3x-6) +5y=43x−6+5

= 4/-6 +5=4−6+5

=4 1/3 or 4.33 (3s.f.)=413or4.33(3s.f.)

xx:
xx-intercept: y=0y=0

4/(3x-6)+5=043x−6+5=0

4/(3x-6) = -543x−6=−5

3x-6 = -0.83x−6=−0.8

3x = -0.8 + 6 = 5.23x=−0.8+6=5.2

x = 5.2/3 = 1.73(3s.f.)x=5.23=1.73(3s.f.)

end behaviour:

y<5y<5 when x<2x<2

e.g. x =1.5x=1.5:

4/(3x-6) +5 = 4/-1.5 +543x−6+5=4−1.5+5

=8/3 + 5=83+5

= 7 2/3=723

y>5y>5 when x>2x>2

e.g. x=2.1x=2.1:

4/(3x-6)+5 = 4/0.3 +543x−6+5=40.3+5

=40/3 + 5=403+5

=18 1/3=1813

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