PROBLEM #2 1. π(π₯) =
3 π₯β4
SOLUTION: ο·
DOMAIN π₯β4=0 π₯=4 = {π₯|π₯πβ\4}
ο·
X-INTERCEPT ππ π β πΌπππΈπ
πΆπΈππ (0,0)
ο·
Y-INTERCEPT 3 3 =β 0β4 4 3 = (0, β ) ππ(0, β0.75) 4
ο·
VERTICAL ASYMPTOTE π₯β4=0 π₯=4
ο·
HORIZONTAL ASYMPTOTE π<π π₯ β ππ₯ππ ππ π¦ = 0
3π₯ 2
2. π(π₯) = π₯ 2 β4
3 =3 1 π¦=3
SOLUTION: ο·
DOMAIN π₯2 β 4 (π₯ β 2)(π₯ + 2) π₯β2=0 π₯=2 π₯+2=0 π₯ = β2 = {π₯|π₯πβ\β2,2}
ο·
X-INTERCEPT ππ π β πΌπππΈπ
πΆπΈππ (0,0)
ο·
Y-INTERCEPT 3(0)2 (0)2 β 4 = (0,0)
ο·
VERTICAL ASYMPTOTE π₯2 β 4 (π₯ β 2)(π₯ + 2) π₯β2=0 π₯=2 π₯+2=0 π₯ = β2 π₯ = β2 πππ π₯ = 2
ο·
HORIZONTAL ASYMPTOTE
3. π(π₯) =
β3π₯ 2 π₯+4
SOLUTION: ο·
DOMAIN π₯+4=0 π₯ = β4 = {π₯|π₯πβ β 4}
ο·
X-INTERCEPT ππ π β πΌπππΈπ
πΆπΈππ (0,0)
ο·
Y-INTERCEPT β3(0)2 =0 0+4 = (0,0)
ο·
VERTICAL ASYMPTOTE π₯+4=0 π₯ = β4
ο·
HORIZONTAL ASYMPTOTE π>π ππ π». π΄.
β3
π₯ = β2 πππ π₯ = 2
4. π(π₯) = π₯ 2 β4 ο·
SOLUTION: ο·
DOMAIN π₯2 β 4 (π₯ β 2)(π₯ + 2) π₯β2=0 π₯=2 π₯+2=0 π₯ = β2 = {π₯|π₯πβ\β2,2}
ο·
X-INTERCEPT ππ π β πΌπππΈπ
πΆπΈππ (0,0)
ο·
Y-INTERCEPT β3 3 = ππ 0.75 2 (0) β 4 4 3 = (0, ) ππ (0,0.75) 4
ο·
VERTICAL ASYMPTOTE π₯2 β 4 (π₯ β 2)(π₯ + 2) π₯β2=0 π₯=2 π₯+2=0 π₯ = β2
HORIZONTAL ASYMPTOTE π<π π¦ β ππ₯ππ ππ π¦ = 0
STUDENT ANSWERED: 1. 2. 3. 4.
A C B D
ACTUAL ANSWERS: 1. 2. 3. 4.
B A D C
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