H Algebra 2/Trig · Unit 4
Find the holes, asymptotes, intercepts, and domain, then sketch the graph.
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Holes, then vertical asymptotes
Factor and cancel first.
A canceled factor is a hole. A denominator zero that remains is a vertical asymptote.
Domain
The domain is all real numbers except the holes and the vertical asymptotes.
Write the excluded values with x ≠.
The other asymptote
Equal degrees give a horizontal asymptote.
A numerator degree that is one more gives an oblique asymptote.
Intercepts and points
X-intercepts are zeros of the reduced numerator. The y-intercept uses x = 0 when it is allowed.
Plot one extra point in each section created by the vertical asymptotes.
No hole
Factor.
(x − 3)(x + 2)
Nothing cancels, so there is no hole.
A vertical asymptote is a denominator zero that does not cancel.
x + 2 = 0
x = −2
The vertical asymptote is x = −2.
The domain is all real numbers, x ≠ −2.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x − 3 = 0
x = 3
The x-intercept is (3, 0).
Substitute x = 0.
y = (0 − 3)(0 + 2)
y = −32
y = −32
The y-intercept is (0, −32).
Choose one point in each section created by the vertical asymptotes.
x = −3
y = (−3 − 3)(−3 + 2)
y = −6−1
y = 6
The point is (−3, 6).
x = −1
y = (−1 − 3)(−1 + 2)
y = −41
y = −4
The point is (−1, −4).
Dashed lines are asymptotes.
Answer: no hole; vertical asymptote x = −2; horizontal asymptote y = 1; domain all real numbers, x ≠ −2; x-intercept (3, 0); y-intercept (0, −32)
Factor.
(x + 1)(x − 4)
Nothing cancels, so there is no hole.
A vertical asymptote is a denominator zero that does not cancel.
x − 4 = 0
x = 4
The vertical asymptote is x = 4.
The domain is all real numbers, x ≠ 4.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x + 1 = 0
x = −1
The x-intercept is (−1, 0).
Substitute x = 0.
y = (0 + 1)(0 − 4)
y = 1−4
y = −14
The y-intercept is (0, −14).
Choose one point in each section created by the vertical asymptotes.
x = 3
y = (3 + 1)(3 − 4)
y = 4−1
y = −4
The point is (3, −4).
x = 5
y = (5 + 1)(5 − 4)
y = 61
y = 6
The point is (5, 6).
Dashed lines are asymptotes.
Answer: no hole; vertical asymptote x = 4; horizontal asymptote y = 1; domain all real numbers, x ≠ 4; x-intercept (−1, 0); y-intercept (0, −14)
Factor.
(x − 5)(x + 3)
Nothing cancels, so there is no hole.
A vertical asymptote is a denominator zero that does not cancel.
x + 3 = 0
x = −3
The vertical asymptote is x = −3.
The domain is all real numbers, x ≠ −3.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x − 5 = 0
x = 5
The x-intercept is (5, 0).
Substitute x = 0.
y = (0 − 5)(0 + 3)
y = −53
y = −53
The y-intercept is (0, −53).
Choose one point in each section created by the vertical asymptotes.
x = −5
y = (−5 − 5)(−5 + 3)
y = −10−2
y = 5
The point is (−5, 5).
x = −2
y = (−2 − 5)(−2 + 3)
y = −71
y = −7
The point is (−2, −7).
Dashed lines are asymptotes.
Answer: no hole; vertical asymptote x = −3; horizontal asymptote y = 1; domain all real numbers, x ≠ −3; x-intercept (5, 0); y-intercept (0, −53)
Factor.
5(x + 1)
Nothing cancels, so there is no hole.
A vertical asymptote is a denominator zero that does not cancel.
x + 1 = 0
x = −1
The vertical asymptote is x = −1.
The domain is all real numbers, x ≠ −1.
The degree of the numerator is less than the degree of the denominator.
y = 0
The horizontal asymptote is y = 0.
The numerator has no zero, so there is no x-intercept.
Substitute x = 0.
y = 5(0 + 1)
y = 51
y = 5
The y-intercept is (0, 5).
Choose one point in each section created by the vertical asymptotes.
x = −2
y = 5(−2 + 1)
y = 5−1
y = −5
The point is (−2, −5).
x = 1
y = 5(1 + 1)
y = 52
The point is (1, 52).
Dashed lines are asymptotes.
Answer: no hole; vertical asymptote x = −1; horizontal asymptote y = 0; domain all real numbers, x ≠ −1; no x-intercept; y-intercept (0, 5)
Factor.
(x − 4)x
Nothing cancels, so there is no hole.
A vertical asymptote is a denominator zero that does not cancel.
x = 0
The vertical asymptote is x = 0.
The domain is all real numbers, x ≠ 0.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x − 4 = 0
x = 4
The x-intercept is (4, 0).
There is no y-intercept because x = 0 is excluded.
Choose one point in each section created by the vertical asymptotes.
x = −1
y = (−1 − 4)(−1)
y = −5−1
y = 5
The point is (−1, 5).
x = 1
y = (1 − 4)(1)
y = −31
y = −3
The point is (1, −3).
Dashed lines are asymptotes.
Answer: no hole; vertical asymptote x = 0; horizontal asymptote y = 1; domain all real numbers, x ≠ 0; x-intercept (4, 0); no y-intercept
Factor.
(x + 4)(x − 1)
Nothing cancels, so there is no hole.
A vertical asymptote is a denominator zero that does not cancel.
x − 1 = 0
x = 1
The vertical asymptote is x = 1.
The domain is all real numbers, x ≠ 1.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x + 4 = 0
x = −4
The x-intercept is (−4, 0).
Substitute x = 0.
y = (0 + 4)(0 − 1)
y = 4−1
y = −4
The y-intercept is (0, −4).
Choose one point in each section created by the vertical asymptotes.
x = −1
y = (−1 + 4)(−1 − 1)
y = 3−2
y = −32
The point is (−1, −32).
x = 2
y = (2 + 4)(2 − 1)
y = 61
y = 6
The point is (2, 6).
Dashed lines are asymptotes.
Answer: no hole; vertical asymptote x = 1; horizontal asymptote y = 1; domain all real numbers, x ≠ 1; x-intercept (−4, 0); y-intercept (0, −4)
A hole and a horizontal asymptote
Factor.
(x − 3)(x + 2)(x − 2)(x + 2)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
(x − 3)(x − 2)
x = −2
Substitute that x-value into the reduced fraction.
y = (−2 − 3)(−2 − 2)
y = −5−4
y = 54
The hole is (−2, 54).
A vertical asymptote is a denominator zero that does not cancel.
x − 2 = 0
x = 2
The vertical asymptote is x = 2.
The domain is all real numbers, x ≠ −2 and x ≠ 2.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x − 3 = 0
x = 3
The x-intercept is (3, 0).
Substitute x = 0.
y = (0 − 3)(0 − 2)
y = −3−2
y = 32
The y-intercept is (0, 32).
Choose one point in each section created by the vertical asymptotes.
x = 1
y = (1 − 3)(1 − 2)
y = −2−1
y = 2
The point is (1, 2).
x = 4
y = (4 − 3)(4 − 2)
y = 12
The point is (4, 12).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (−2, 54); vertical asymptote x = 2; horizontal asymptote y = 1; domain all real numbers, x ≠ −2 and x ≠ 2; x-intercept (3, 0); y-intercept (0, 32)
Factor.
(x + 4)(x − 3)(x − 3)(x + 3)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
(x + 4)(x + 3)
x = 3
Substitute that x-value into the reduced fraction.
y = (3 + 4)(3 + 3)
y = 76
The hole is (3, 76).
A vertical asymptote is a denominator zero that does not cancel.
x + 3 = 0
x = −3
The vertical asymptote is x = −3.
The domain is all real numbers, x ≠ −3 and x ≠ 3.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x + 4 = 0
x = −4
The x-intercept is (−4, 0).
Substitute x = 0.
y = (0 + 4)(0 + 3)
y = 43
The y-intercept is (0, 43).
Choose one point in each section created by the vertical asymptotes.
x = −5
y = (−5 + 4)(−5 + 3)
y = −1−2
y = 12
The point is (−5, 12).
x = −2
y = (−2 + 4)(−2 + 3)
y = 21
y = 2
The point is (−2, 2).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (3, 76); vertical asymptote x = −3; horizontal asymptote y = 1; domain all real numbers, x ≠ −3 and x ≠ 3; x-intercept (−4, 0); y-intercept (0, 43)
Factor.
(x − 4)(x + 3)(x − 4)(x + 4)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
(x + 3)(x + 4)
x = 4
Substitute that x-value into the reduced fraction.
y = (4 + 3)(4 + 4)
y = 78
The hole is (4, 78).
A vertical asymptote is a denominator zero that does not cancel.
x + 4 = 0
x = −4
The vertical asymptote is x = −4.
The domain is all real numbers, x ≠ −4 and x ≠ 4.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x + 3 = 0
x = −3
The x-intercept is (−3, 0).
Substitute x = 0.
y = (0 + 3)(0 + 4)
y = 34
The y-intercept is (0, 34).
Choose one point in each section created by the vertical asymptotes.
x = −5
y = (−5 + 3)(−5 + 4)
y = −2−1
y = 2
The point is (−5, 2).
x = −2
y = (−2 + 3)(−2 + 4)
y = 12
The point is (−2, 12).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (4, 78); vertical asymptote x = −4; horizontal asymptote y = 1; domain all real numbers, x ≠ −4 and x ≠ 4; x-intercept (−3, 0); y-intercept (0, 34)
Factor.
(x + 1)(x + 2)(x − 1)(x + 1)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
(x + 2)(x − 1)
x = −1
Substitute that x-value into the reduced fraction.
y = (−1 + 2)(−1 − 1)
y = 1−2
y = −12
The hole is (−1, −12).
A vertical asymptote is a denominator zero that does not cancel.
x − 1 = 0
x = 1
The vertical asymptote is x = 1.
The domain is all real numbers, x ≠ −1 and x ≠ 1.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x + 2 = 0
x = −2
The x-intercept is (−2, 0).
Substitute x = 0.
y = (0 + 2)(0 − 1)
y = 2−1
y = −2
The y-intercept is (0, −2).
Choose one point in each section created by the vertical asymptotes.
x = −3
y = (−3 + 2)(−3 − 1)
y = −1−4
y = 14
The point is (−3, 14).
x = 2
y = (2 + 2)(2 − 1)
y = 41
y = 4
The point is (2, 4).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (−1, −12); vertical asymptote x = 1; horizontal asymptote y = 1; domain all real numbers, x ≠ −1 and x ≠ 1; x-intercept (−2, 0); y-intercept (0, −2)
Factor.
(x − 5)(x + 1)(x − 5)(x + 5)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
(x + 1)(x + 5)
x = 5
Substitute that x-value into the reduced fraction.
y = (5 + 1)(5 + 5)
y = 610
y = 35
The hole is (5, 35).
A vertical asymptote is a denominator zero that does not cancel.
x + 5 = 0
x = −5
The vertical asymptote is x = −5.
The domain is all real numbers, x ≠ −5 and x ≠ 5.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x + 1 = 0
x = −1
The x-intercept is (−1, 0).
Substitute x = 0.
y = (0 + 1)(0 + 5)
y = 15
The y-intercept is (0, 15).
Choose one point in each section created by the vertical asymptotes.
x = −6
y = (−6 + 1)(−6 + 5)
y = −5−1
y = 5
The point is (−6, 5).
x = −4
y = (−4 + 1)(−4 + 5)
y = −31
y = −3
The point is (−4, −3).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (5, 35); vertical asymptote x = −5; horizontal asymptote y = 1; domain all real numbers, x ≠ −5 and x ≠ 5; x-intercept (−1, 0); y-intercept (0, 15)
Factor.
(x − 1)(x − 3)(x − 1)(x + 1)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
(x − 3)(x + 1)
x = 1
Substitute that x-value into the reduced fraction.
y = (1 − 3)(1 + 1)
y = −22
y = −1
The hole is (1, −1).
A vertical asymptote is a denominator zero that does not cancel.
x + 1 = 0
x = −1
The vertical asymptote is x = −1.
The domain is all real numbers, x ≠ −1 and x ≠ 1.
The degrees are equal, so the horizontal asymptote is the ratio of the leading coefficients.
y = 1
The horizontal asymptote is y = 1.
An x-intercept is a zero of the reduced numerator.
x − 3 = 0
x = 3
The x-intercept is (3, 0).
Substitute x = 0.
y = (0 − 3)(0 + 1)
y = −31
y = −3
The y-intercept is (0, −3).
Choose one point in each section created by the vertical asymptotes.
x = −2
y = (−2 − 3)(−2 + 1)
y = −5−1
y = 5
The point is (−2, 5).
x = 2
y = (2 − 3)(2 + 1)
y = −13
y = −13
The point is (2, −13).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (1, −1); vertical asymptote x = −1; horizontal asymptote y = 1; domain all real numbers, x ≠ −1 and x ≠ 1; x-intercept (3, 0); y-intercept (0, −3)
A hole and an oblique asymptote
Factor.
x(x − 2)(x + 2)(x + 3)(x − 2)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
x(x + 2)(x + 3)
x = 2
Substitute that x-value into the reduced fraction.
y = (2)(2 + 2)(2 + 3)
y = 85
The hole is (2, 85).
A vertical asymptote is a denominator zero that does not cancel.
x + 3 = 0
x = −3
The vertical asymptote is x = −3.
The domain is all real numbers, x ≠ −3 and x ≠ 2.
The degree of the numerator is one more than the degree of the denominator, so the other asymptote is oblique.
(x + 3)(x − 1) = x² + 2x − 3
x² + 2x = (x² + 2x − 3) + 3
y = x − 1
The oblique asymptote is y = x − 1.
An x-intercept is a zero of the reduced numerator.
x + 2 = 0
x = −2
The x-intercept is (−2, 0).
x = 0
The x-intercept is (0, 0).
Substitute x = 0.
y = (0)(0 + 2)(0 + 3)
y = 03
y = 0
The y-intercept is (0, 0).
Choose one point in each section created by the vertical asymptotes.
x = −4
y = (−4)(−4 + 2)(−4 + 3)
y = 8−1
y = −8
The point is (−4, −8).
x = −1
y = (−1)(−1 + 2)(−1 + 3)
y = −12
y = −12
The point is (−1, −12).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (2, 85); vertical asymptote x = −3; oblique asymptote y = x − 1; domain all real numbers, x ≠ −3 and x ≠ 2; x-intercept (−2, 0), (0, 0); y-intercept (0, 0)
Factor.
x(x − 3)(x + 3)(x − 3)(x + 2)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
x(x + 3)(x + 2)
x = 3
Substitute that x-value into the reduced fraction.
y = (3)(3 + 3)(3 + 2)
y = 185
The hole is (3, 185).
A vertical asymptote is a denominator zero that does not cancel.
x + 2 = 0
x = −2
The vertical asymptote is x = −2.
The domain is all real numbers, x ≠ −2 and x ≠ 3.
The degree of the numerator is one more than the degree of the denominator, so the other asymptote is oblique.
(x + 2)(x + 1) = x² + 3x + 2
x² + 3x = (x² + 3x + 2) − 2
y = x + 1
The oblique asymptote is y = x + 1.
An x-intercept is a zero of the reduced numerator.
x + 3 = 0
x = −3
The x-intercept is (−3, 0).
x = 0
The x-intercept is (0, 0).
Substitute x = 0.
y = (0)(0 + 3)(0 + 2)
y = 02
y = 0
The y-intercept is (0, 0).
Choose one point in each section created by the vertical asymptotes.
x = −4
y = (−4)(−4 + 3)(−4 + 2)
y = 4−2
y = −2
The point is (−4, −2).
x = −1
y = (−1)(−1 + 3)(−1 + 2)
y = −21
y = −2
The point is (−1, −2).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (3, 185); vertical asymptote x = −2; oblique asymptote y = x + 1; domain all real numbers, x ≠ −2 and x ≠ 3; x-intercept (−3, 0), (0, 0); y-intercept (0, 0)
Factor.
x(x + 2)(x − 1)(x − 2)(x + 2)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
x(x − 1)(x − 2)
x = −2
Substitute that x-value into the reduced fraction.
y = (−2)(−2 − 1)(−2 − 2)
y = 6−4
y = −32
The hole is (−2, −32).
A vertical asymptote is a denominator zero that does not cancel.
x − 2 = 0
x = 2
The vertical asymptote is x = 2.
The domain is all real numbers, x ≠ −2 and x ≠ 2.
The degree of the numerator is one more than the degree of the denominator, so the other asymptote is oblique.
(x − 2)(x + 1) = x² − x − 2
x² − x = (x² − x − 2) + 2
y = x + 1
The oblique asymptote is y = x + 1.
An x-intercept is a zero of the reduced numerator.
x = 0
The x-intercept is (0, 0).
x − 1 = 0
x = 1
The x-intercept is (1, 0).
Substitute x = 0.
y = (0)(0 − 1)(0 − 2)
y = 0−2
y = 0
The y-intercept is (0, 0).
Choose one point in each section created by the vertical asymptotes.
x = −1
y = (−1)(−1 − 1)(−1 − 2)
y = 2−3
y = −23
The point is (−1, −23).
x = 3
y = (3)(3 − 1)(3 − 2)
y = 61
y = 6
The point is (3, 6).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (−2, −32); vertical asymptote x = 2; oblique asymptote y = x + 1; domain all real numbers, x ≠ −2 and x ≠ 2; x-intercept (0, 0), (1, 0); y-intercept (0, 0)
Factor.
2x(x − 2)(x + 2)(x − 2)(x + 1)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
2x(x + 2)(x + 1)
x = 2
Substitute that x-value into the reduced fraction.
y = 2(2)(2 + 2)(2 + 1)
y = 163
The hole is (2, 163).
A vertical asymptote is a denominator zero that does not cancel.
x + 1 = 0
x = −1
The vertical asymptote is x = −1.
The domain is all real numbers, x ≠ −1 and x ≠ 2.
The degree of the numerator is one more than the degree of the denominator, so the other asymptote is oblique.
(x + 1)(2x + 2) = 2x² + 4x + 2
2x² + 4x = (2x² + 4x + 2) − 2
y = 2x + 2
The oblique asymptote is y = 2x + 2.
An x-intercept is a zero of the reduced numerator.
x + 2 = 0
x = −2
The x-intercept is (−2, 0).
x = 0
The x-intercept is (0, 0).
Substitute x = 0.
y = 2(0)(0 + 2)(0 + 1)
y = 01
y = 0
The y-intercept is (0, 0).
Choose one point in each section created by the vertical asymptotes.
x = −3
y = 2(−3)(−3 + 2)(−3 + 1)
y = 6−2
y = −3
The point is (−3, −3).
x = 1
y = 2(1)(1 + 2)(1 + 1)
y = 62
y = 3
The point is (1, 3).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (2, 163); vertical asymptote x = −1; oblique asymptote y = 2x + 2; domain all real numbers, x ≠ −1 and x ≠ 2; x-intercept (−2, 0), (0, 0); y-intercept (0, 0)
Factor.
x(x − 1)(x − 3)(x − 1)(x − 4)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
x(x − 3)(x − 4)
x = 1
Substitute that x-value into the reduced fraction.
y = (1)(1 − 3)(1 − 4)
y = −2−3
y = 23
The hole is (1, 23).
A vertical asymptote is a denominator zero that does not cancel.
x − 4 = 0
x = 4
The vertical asymptote is x = 4.
The domain is all real numbers, x ≠ 1 and x ≠ 4.
The degree of the numerator is one more than the degree of the denominator, so the other asymptote is oblique.
(x − 4)(x + 1) = x² − 3x − 4
x² − 3x = (x² − 3x − 4) + 4
y = x + 1
The oblique asymptote is y = x + 1.
An x-intercept is a zero of the reduced numerator.
x = 0
The x-intercept is (0, 0).
x − 3 = 0
x = 3
The x-intercept is (3, 0).
Substitute x = 0.
y = (0)(0 − 3)(0 − 4)
y = 0−4
y = 0
The y-intercept is (0, 0).
Choose one point in each section created by the vertical asymptotes.
x = 2
y = (2)(2 − 3)(2 − 4)
y = −2−2
y = 1
The point is (2, 1).
x = 6
y = (6)(6 − 3)(6 − 4)
y = 182
y = 9
The point is (6, 9).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (1, 23); vertical asymptote x = 4; oblique asymptote y = x + 1; domain all real numbers, x ≠ 1 and x ≠ 4; x-intercept (0, 0), (3, 0); y-intercept (0, 0)
Factor.
x(x + 4)(x + 1)(x + 1)(x + 2)
Cancel the common factor. A canceled factor is a hole, not a vertical asymptote.
x(x + 4)(x + 2)
x = −1
Substitute that x-value into the reduced fraction.
y = (−1)(−1 + 4)(−1 + 2)
y = −31
y = −3
The hole is (−1, −3).
A vertical asymptote is a denominator zero that does not cancel.
x + 2 = 0
x = −2
The vertical asymptote is x = −2.
The domain is all real numbers, x ≠ −2 and x ≠ −1.
The degree of the numerator is one more than the degree of the denominator, so the other asymptote is oblique.
(x + 2)(x + 2) = x² + 4x + 4
x² + 4x = (x² + 4x + 4) − 4
y = x + 2
The oblique asymptote is y = x + 2.
An x-intercept is a zero of the reduced numerator.
x + 4 = 0
x = −4
The x-intercept is (−4, 0).
x = 0
The x-intercept is (0, 0).
Substitute x = 0.
y = (0)(0 + 4)(0 + 2)
y = 02
y = 0
The y-intercept is (0, 0).
Choose one point in each section created by the vertical asymptotes.
x = −3
y = (−3)(−3 + 4)(−3 + 2)
y = −3−1
y = 3
The point is (−3, 3).
x = 1
y = (1)(1 + 4)(1 + 2)
y = 53
The point is (1, 53).
The open circle is a hole. Dashed lines are asymptotes.
Answer: hole (−1, −3); vertical asymptote x = −2; oblique asymptote y = x + 2; domain all real numbers, x ≠ −2 and x ≠ −1; x-intercept (−4, 0), (0, 0); y-intercept (0, 0)
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