Try It
The path passes through the origin and has vertex at ( −4,7 ), ( −4,7 ), so h(x)=– 7 16 (x+4) 2 +7. h(x)=– 7 16 (x+4) 2 +7. To make the shot, h( −7.5 ) h( −7.5 ) would need to be about 4 but h(–7.5)≈1.64; h(–7.5)≈1.64; he doesn’t make it.
g(x)= x 2 −6x+13 g(x)= x 2 −6x+13 in general form; g(x)= (x−3) 2 +4 g(x)= (x−3) 2 +4 in standard form
The domain is all real numbers. The range is f(x)≥ 8 11 , f(x)≥ 8 11 , or [ 8 11 ,∞ ). [ 8 11 ,∞ ).
y-intercept at (0, 13), No x- x- intercepts
ⓐ3 seconds
ⓑ256 feet
ⓒ7 seconds
f(x) f(x) is a power function because it can be written as f(x)=8 x 4 . f(x)=8 x 4 . The other functions are not power functions.
As x x approaches positive or negative infinity, f( x ) f( x ) decreases without bound: as x→±∞, f(x)→−∞ x→±∞, f(x)→−∞ because of the negative coefficient.
The degree is 6. The leading term is − x 6 . − x 6 . The leading coefficient is −1. −1.
As x→∞, f(x)→−∞; as x→−∞, f(x)→−∞. x→∞, f(x)→−∞; as x→−∞, f(x)→−∞. It has the shape of an even degree power function with a negative coefficient.
The leading term is 0.2 x 3 , 0.2 x 3 , so it is a degree 3 polynomial. As x x approaches positive infinity, f( x ) f( x ) increases without bound; as x x approaches negative infinity, f( x ) f( x ) decreases without bound.
y-intercept (0,0); (0,0); x-intercepts (0,0),(–2,0), (0,0),(–2,0), and (5,0) (5,0)
There are at most 12 x- x- intercepts and at most 11 turning points.
The end behavior indicates an odd-degree polynomial function; there are 3 x- x- intercepts and 2 turning points, so the degree is odd and at least 3. Because of the end behavior, we know that the lead coefficient must be negative.
The x- x- intercepts are (2,0),(−1,0), (2,0),(−1,0), and (5,0), (5,0), the y-intercept is (0,2), (0,2), and the graph has at most 2 turning points.
y-intercept (0,0); (0,0); x-intercepts (0,0),(–5,0),(2,0), (0,0),(–5,0),(2,0), and (3,0) (3,0)
The graph has a zero of –5 with multiplicity 3, a zero of -1 with multiplicity 2, and a zero of 3 with multiplicity 4.
Because f f is a polynomial function and since f(1) f(1) is negative and f(2) f(2) is positive, there is at least one real zero between x=1 x=1 and x=2. x=2.
f(x)=− 1 8 (x−2) 3 (x+1) 2 (x−4) f(x)=− 1 8 (x−2) 3 (x+1) 2 (x−4)
The minimum occurs at approximately the point (0,−6.5), (0,−6.5), and the maximum occurs at approximately the point (3.5,7). (3.5,7).
4 x 2 −8x+15− 78 4x+5 4 x 2 −8x+15− 78 4x+5
3 x 3 −3 x 2 +21x−150+ 1,090 x+7 3 x 3 −3 x 2 +21x−150+ 1,090 x+7
3 x 2 −4x+1 3 x 2 −4x+1
f(−3)=−412 f(−3)=−412
The zeros are 2, –2, and –4.
There are no rational zeros.
The zeros are –4, 1 2 ,and 1. –4, 1 2 ,and 1.
f(x)=− 1 2 x 3 + 5 2 x 2 −2x+10 f(x)=− 1 2 x 3 + 5 2 x 2 −2x+10
There must be 4, 2, or 0 positive real roots and 0 negative real roots. The graph shows that there are 2 positive real zeros and 0 negative real zeros.
3 meters by 4 meters by 7 meters
End behavior: as x→±∞, f(x)→0; x→±∞, f(x)→0; Local behavior: as x→0, f(x)→∞ x→0, f(x)→∞ (there are no x- or y-intercepts)
The function and the asymptotes are shifted 3 units right and 4 units down. As x→3,f(x)→∞, x→3,f(x)→∞, and as x→±∞,f(x)→−4. x→±∞,f(x)→−4.
The function is f(x)= 1 (x−3) 2 −4. f(x)= 1 (x−3) 2 −4.
12 11 12 11
The domain is all real numbers except x=1 x=1 and x=5. x=5.
Removable discontinuity at x=5. x=5. Vertical asymptotes: x=0,x=1. x=0,x=1.
Vertical asymptotes at x=2 x=2 and x=–3; x=–3; horizontal asymptote at y=4. y=4.
For the transformed reciprocal squared function, we find the rational form. f(x)= 1 (x−3) 2 −4= 1−4 (x−3) 2 (x−3) 2 = 1−4( x 2 −6x+9) (x−3)(x−3) = −4 x 2 +24x−35 x 2 −6x+9 f(x)= 1 (x−3) 2 −4= 1−4 (x−3) 2 (x−3) 2 = 1−4( x 2 −6x+9) (x−3)(x−3) = −4 x 2 +24x−35 x 2 −6x+9
Because the numerator is the same degree as the denominator we know that as x→±∞, f(x)→−4; so y=–4 x→±∞, f(x)→−4; so y=–4 is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is x=3, x=3, because as x→3,f(x)→∞. x→3,f(x)→∞. We then set the numerator equal to 0 and find the x-intercepts are at (2.5,0) (2.5,0) and (3.5,0). (3.5,0). Finally, we evaluate the function at 0 and find the y-intercept to be at ( 0, −35 9 ). ( 0, −35 9 ).
Horizontal asymptote at y= 1 2 . y= 1 2 . Vertical asymptotes at x=1 and x=3. x=1 and x=3. y-intercept at ( 0, 4 3 . ) ( 0, 4 3 . )
x-intercepts at (2,0) and (–2,0). (2,0) and (–2,0). (–2,0) (–2,0) is a zero with multiplicity 2, and the graph bounces off the x-axis at this point. (2,0) (2,0) is a single zero and the graph crosses the axis at this point.
f −1 ( f( x ) )= f −1 ( x+5 3 )=3( x+5 3 )−5=( x−5 )+5=x f −1 ( f( x ) )= f −1 ( x+5 3 )=3( x+5 3 )−5=( x−5 )+5=x and f( f −1 ( x ) )=f( 3x−5 )= ( 3x−5 )+5 3 = 3x 3 =x f( f −1 ( x ) )=f( 3x−5 )= ( 3x−5 )+5 3 = 3x 3 =x
f −1 (x)= x 3 −4 f −1 (x)= x 3 −4
f −1 (x)= x−1 f −1 (x)= x−1
f −1 (x)= x 2 −3 2 ,x≥0 f −1 (x)= x 2 −3 2 ,x≥0
f −1 (x)= 2x+3 x−1 f −1 (x)= 2x+3 x−1
128 3 128 3
9 2 9 2
x=20 x=20
5.1 Section Exercises
When written in that form, the vertex can be easily identified.
If a=0 a=0 then the function becomes a linear function.
If possible, we can use factoring. Otherwise, we can use the quadratic formula.
g(x)= (x+1) 2 −4, g(x)= (x+1) 2 −4, Vertex ( −1,−4 ) ( −1,−4 )
f(x)= ( x+ 5 2 ) 2 − 33 4 , f(x)= ( x+ 5 2 ) 2 − 33 4 , Vertex ( − 5 2 ,− 33 4 ) ( − 5 2 ,− 33 4 )
f(x)=3 (x−1) 2 −12, f(x)=3 (x−1) 2 −12, Vertex (1,−12) (1,−12)
f(x)=3 ( x− 5 6 ) 2 − 37 12 , f(x)=3 ( x− 5 6 ) 2 − 37 12 , Vertex ( 5 6 ,− 37 12 ) ( 5 6 ,− 37 12 )
Minimum is − 17 2 − 17 2 and occurs at 5 2 . 5 2 . Axis of symmetry is x= 5 2 . x= 5 2 .
Minimum is − 17 16 − 17 16 and occurs at − 1 8 . − 1 8 . Axis of symmetry is x=− 1 8 . x=− 1 8 .
Minimum is − 7 2 − 7 2 and occurs at −3. −3. Axis of symmetry is x=−3. x=−3.
Domain is ( −∞,∞ ). ( −∞,∞ ). Range is [2,∞). [2,∞).
Domain is ( −∞,∞ ). ( −∞,∞ ). Range is [−5,∞). [−5,∞).
Domain is ( −∞,∞ ). ( −∞,∞ ). Range is [−12,∞). [−12,∞).
f(x)= x 2 +4x+3 f(x)= x 2 +4x+3
f(x)= x 2 -4x+7 f(x)= x 2 -4x+7
f(x)= -149 x 2 +649x +8949 f(x)= -149 x 2 +649x +8949
f(x)= x 2 -2x+1 f(x)= x 2 -2x+1
Vertex: (3, −10), axis of symmetry: x = 3, intercepts: (3+10,0)(3+10,0) and (3-10,0)(3-10,0)
Vertex: ( 7 2 , − 37 4 ) ( 7 2 , − 37 4 ), axis of symmetry: x=72 x=72, y-intercept: (0,3) (0,3), x-intercepts: ( 7+37 2 , 0 ), ( 7−37 2 , 0 ) ( 7+37 2 , 0 ), ( 7−37 2 , 0 )
Vertex: (32,-12) (32,-12), axis of symmetry: x=32 x=32, intercept: ( 3+23 2 , 0) ( 3+23 2 , 0) and ( 3-23 2 , 0) ( 3-23 2 , 0)
f(x)= x 2 +2x+3
f(x)=-3 x 2 −6x−1
f(x)=-14 x 2 −x+2
f(x)= x 2 +2x+1
f(x)= - x 2 +2x
f(x)=2 x 2
ਗ੍ਰਾਫ ਸੱਜੇ ਜਾਂ ਖੱਬੇ ਪਾਸੇ ਖਿਸਕ ਜਾਂਦਾ ਹੈ (ਖਿਤਿਜੀ ਖਿਸਕਾਅ)।
ਸਸਪੈਂਸ਼ਨ ਬ੍ਰਿਜ ਦਾ ਕੇਂਦਰ ਤੋਂ 1,000 ਫੁੱਟ ਦਾ ਫਾਸਲਾ ਹੈ।
ਡੋਮੇਨ (−∞,∞) ਹੈ। ਰੇਂਜ (-∞,2] ਹੈ।
ਡੋਮੇਨ: (-∞,∞) ; ਰੇਂਜ: [100,∞)
f(x)=2 x 2 +2
f(x)=- x 2 −2
f(x)=3 x 2 +6x-15
75 ਫੁੱਟ ਗੁਣਾ 50 ਫੁੱਟ
3 ਅਤੇ 3; ਗੁਣਨਫਲ 9 ਹੈ
1800 ਹਜ਼ਾਰ ਫੋਨਾਂ ਦੇ ਉਤਪਾਦਨ 'ਤੇ ਮਾਲੀਆ ਵੱਧ ਤੋਂ ਵੱਧ ਮੁੱਲ ਤੱਕ ਪਹੁੰਚਦਾ ਹੈ।
2.449 ਸੈਕਿੰਡ
41 ਰੁੱਖ ਪ੍ਰਤੀ ਏਕੜ
5.2 ਸੈਕਸ਼ਨ ਅਭਿਆਸ
ਪਾਵਰ ਫੰਕਸ਼ਨ ਦਾ ਗੁਣਾਂਕ ਉਹ ਅਸਲ ਸੰਖਿਆ ਹੈ ਜੋ ਕਿਸੇ ਘਾਤ ਤੱਕ ਵਧਾਈ ਗਈ ਚੱਲ ਨਾਲ ਗੁਣਾ ਹੁੰਦੀ ਹੈ। ਡਿਗਰੀ ਫੰਕਸ਼ਨ ਵਿੱਚ ਆਉਣ ਵਾਲੀ ਸਭ ਤੋਂ ਵੱਡੀ ਘਾਤ ਹੈ।
ਜਿਵੇਂ x x ਬੇਅੰਤ ਘੱਟਦਾ ਹੈ, ਉਸੇ ਤਰ੍ਹਾਂ f( x ) ਵੀ ਘੱਟਦਾ ਹੈ। ਜਿਵੇਂ x x ਬੇਅੰਤ ਵਧਦਾ ਹੈ, ਉਸੇ ਤਰ੍ਹਾਂ f( x ) ਵੀ ਵਧਦਾ ਹੈ।
ਪੋਲੀਨੋਮੀਅਲ ਫੰਕਸ਼ਨ ਸਮ ਡਿਗਰੀ ਦਾ ਹੈ ਅਤੇ ਲੀਡਿੰਗ ਗੁਣਾਂਕ ਨਕਾਰਾਤਮਕ ਹੈ।
ਪਾਵਰ ਫੰਕਸ਼ਨ
ਨਾ ਤਾਂ
1. ਨਾ ਹੀ
2. ਘਾਤ = 2, ਗੁਣਾਂਕ = –2
3. ਘਾਤ = 4, ਗੁਣਾਂਕ = –2
4. ਜਿਉਂ ਜਿਉਂ x → ∞, f(x) → ∞, ਜਿਉਂ ਜਿਉਂ x → −∞, f(x) → ∞
5. ਜਿਉਂ ਜਿਉਂ x → −∞, f(x) → −∞, ਜਿਉਂ ਜਿਉਂ x → ∞, f(x) → −∞
6. ਜਿਉਂ ਜਿਉਂ x → −∞, f(x) → −∞, ਜਿਉਂ ਜਿਉਂ x → ∞, f(x) → −∞
7. ਜਿਉਂ ਜਿਉਂ x → ∞, f(x) → ∞, ਜਿਉਂ ਜਿਉਂ x → −∞, f(x) → −∞
8. y-ਅੰਤਰਖੰਡ (0,12) ਹੈ, t-ਅੰਤਰਖੰਡ (1,0); (–2,0); ਅਤੇ (3,0) ਹਨ।
9. y-ਅੰਤਰਖੰਡ (0,−16) ਹੈ। x-ਅੰਤਰਖੰਡ (2,0) ਅਤੇ (−2,0) ਹਨ।
10. y-ਅੰਤਰਖੰਡ (0,0) ਹੈ। x-ਅੰਤਰਖੰਡ (0,0), (4,0), ਅਤੇ ( −2, 0 ) ਹਨ।
11. 3
12. 5
13. 3
14. 5
15. ਹਾਂ। ਮੁੜਨ ਬਿੰਦੂਆਂ ਦੀ ਗਿਣਤੀ 2 ਹੈ। ਘੱਟੋ-ਘੱਟ ਸੰਭਵ ਘਾਤ 3 ਹੈ।
16. ਹਾਂ। ਮੁੜਨ ਬਿੰਦੂਆਂ ਦੀ ਗਿਣਤੀ 1 ਹੈ। ਘੱਟੋ-ਘੱਟ ਸੰਭਵ ਘਾਤ 2 ਹੈ।
17. ਹਾਂ। ਮੁੜਨ ਬਿੰਦੂਆਂ ਦੀ ਗਿਣਤੀ 0 ਹੈ। ਘੱਟੋ-ਘੱਟ ਸੰਭਵ ਘਾਤ 1 ਹੈ।
18. ਹਾਂ। ਮੁੜਨ ਬਿੰਦੂਆਂ ਦੀ ਗਿਣਤੀ 0 ਹੈ। ਘੱਟੋ-ਘੱਟ ਸੰਭਵ ਘਾਤ 1 ਹੈ।
19. ਕਤਾਰ: x | f( x )
20. ਕਤਾਰ: 10 | 9,500
21. ਕਤਾਰ: 100 | 99,950,000
22. ਕਤਾਰ: –10 | 9,500
23. ਕਤਾਰ: –100 | 99,950,000
24. ਜਿਉਂ ਜਿਉਂ x → −∞, f(x) → ∞, ਜਿਉਂ ਜਿਉਂ x → ∞, f(x) → ∞
row: x x | f( x ) f( x )
row: 10 | –504
row: 100 | –941,094
row: –10 | 1,716
row: –100 | 1,061,106
As x→−∞ x→−∞, f(x)→∞,asx→∞,f(x)→−∞ f(x)→∞,asx→∞,f(x)→−∞
The y- y- intercept is ( 0, 0 ). ( 0, 0 ). The x- x- intercepts are ( 0, 0 ),( 2, 0 ). ( 0, 0 ),( 2, 0 ). As x→−∞ x→−∞, f(x)→∞,asx→∞,f(x)→∞ f(x)→∞,asx→∞,f(x)→∞
The y- y- intercept is ( 0,0 ) ( 0,0 ) . The x- x- intercepts are ( 0, 0 ),( 5, 0 ),( 7, 0 ). ( 0, 0 ),( 5, 0 ),( 7, 0 ). As x→−∞ x→−∞, f(x)→−∞,asx→∞,f(x)→∞ f(x)→−∞,asx→∞,f(x)→∞
The y- y- intercept is ( 0, 0 ). ( 0, 0 ). The x- x- intercept is ( −4, 0 ),( 0, 0 ),( 4, 0 ). ( −4, 0 ),( 0, 0 ),( 4, 0 ). Asx→−∞ Asx→−∞, f(x)→−∞,asx→∞,f(x)→∞ f(x)→−∞,asx→∞,f(x)→∞
The y- y- intercept is ( 0, −81 ). ( 0, −81 ). The x- x- intercept are ( 3, 0 ),( −3, 0 ). ( 3, 0 ),( −3, 0 ). As x→−∞ x→−∞, f(x)→∞,asx→∞,f(x)→∞ f(x)→∞,asx→∞,f(x)→∞
The y- y- intercept is ( 0, 0 ). ( 0, 0 ). The x- x- intercepts are ( −3, 0 ),( 0, 0 ),( 5, 0 ). ( −3, 0 ),( 0, 0 ),( 5, 0 ). As x→−∞ x→−∞, f(x)→−∞,asx→∞,f(x)→∞ f(x)→−∞,asx→∞,f(x)→∞
f(x)= x 2 −4 f(x)= x 2 −4
f(x)= x 3 −4 x 2 +4x f(x)= x 3 −4 x 2 +4x
f(x)= x 4 +1 f(x)= x 4 +1
V(m)=8 m 3 +36 m 2 +54m+27 V(m)=8 m 3 +36 m 2 +54m+27
V(x)=4 x 3 −32 x 2 +64x V(x)=4 x 3 −32 x 2 +64x
5.3 Section Exercises
The x- x- intercept is where the graph of the function crosses the x- x- axis, and the zero of the function is the input value for which f(x)=0. f(x)=0.
If we evaluate the function at a a and at b b and the sign of the function value changes, then we know a zero exists between a a and b. b.
There will be a factor raised to an even power.
(−2,0),(3,0),(−5,0) (−2,0),(3,0),(−5,0)
(3,0),(−1,0),(0,0) (3,0),(−1,0),(0,0)
( 0,0 ),( −5,0 ),( 2,0 ) ( 0,0 ),( −5,0 ),( 2,0 )
( 0,0 ),( −5,0 ),( 4,0 ) ( 0,0 ),( −5,0 ),( 4,0 )
(2,0), (−2,0), (−1,0) (2,0), (−2,0), (−1,0)
(−2,0),(2,0),(1/2,0) (−2,0),(2,0),(1/2,0)
(1,0), (−1,0) (1,0), (−1,0)
(0,0), (√3,0), (−√3,0) (0,0), (√3,0), (−√3,0)
(0,0), (1,0), (−1,0), (2,0), (−2,0) (0,0), (1,0), (−1,0), (2,0), (−2,0)
f(2)=–10 f(2)=–10 ਅਤੇ f(4)=28। f(4)=28। ਚਿੰਨ੍ਹ ਬਦਲਾਵ ਦੀ ਪੁਸ਼ਟੀ ਕਰਦਾ ਹੈ।
f(1)=3 f(1)=3 ਅਤੇ f(3)=–77। f(3)=–77। ਚਿੰਨ੍ਹ ਬਦਲਾਵ ਦੀ ਪੁਸ਼ਟੀ ਕਰਦਾ ਹੈ।
f(0.01)=1.000001 f(0.01)=1.000001 ਅਤੇ f(0.1)=–7.999। f(0.1)=–7.999। ਚਿੰਨ੍ਹ ਬਦਲਾਵ ਦੀ ਪੁਸ਼ਟੀ ਕਰਦਾ ਹੈ।
0 ਦੋਹਰੀ ਬਹੁਲਤਾ ਨਾਲ, −3/2 ਪੰਜ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ, 4 ਦੋਹਰੀ ਬਹੁਲਤਾ ਨਾਲ
0 ਦੋਹਰੀ ਬਹੁਲਤਾ ਨਾਲ, –2 ਦੋਹਰੀ ਬਹੁਲਤਾ ਨਾਲ
−2/3 ਪੰਜ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ, 5 ਦੋਹਰੀ ਬਹੁਲਤਾ ਨਾਲ
0 ਚਾਰ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ, 2 ਇੱਕ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ, −1 ਇੱਕ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ
3/2 ਦੋਹਰੀ ਬਹੁਲਤਾ ਨਾਲ, 0 ਤਿੰਨ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ
0 ਛੇ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ, 2/3 ਦੋਹਰੀ ਬਹੁਲਤਾ ਨਾਲ
x-ਅੰਤਰਖੰਡ, (1,0) ਦੋਹਰੀ ਬਹੁਲਤਾ ਨਾਲ, (–4,0) ਇੱਕ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ, y-ਅੰਤਰਖੰਡ (0,4)। ਜਿਵੇਂ x→−∞, f(x)→−∞, ਜਿਵੇਂ x→∞, f(x)→∞।
x-ਅੰਤਰਖੰਡ (3,0) ਤਿੰਨ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ, (2,0) ਦੋਹਰੀ ਬਹੁਲਤਾ ਨਾਲ, y-ਅੰਤਰਖੰਡ (0,–108)। ਜਿਵੇਂ x→−∞, f(x)→−∞, ਜਿਵੇਂ x→∞, f(x)→∞।
x-ਅੰਤਰਖੰਡ (0,0), (–2,0), (4,0) ਇੱਕ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ, y-ਅੰਤਰਖੰਡ (0,0)। ਜਿਵੇਂ x→−∞, f(x)→∞, ਜਿਵੇਂ x→∞, f(x)→−∞।
f(x)=−2/9 (x−3)(x+1)(x+3)
f(x)=1/4 (x+2)² (x−3)
–4, –2, 1, 3 ਇੱਕ ਗੁਣਾ ਬਹੁਲਤਾ ਨਾਲ
–2, 3 ਹਰੇਕ ਦੋਹਰੀ ਬਹੁਲਤਾ ਨਾਲ
f(x)=−2/3 (x+2)(x−1)(x−3)
f(x)=1/3 (x−3)² (x−1)² (x+3)
f(x)=−15 (x−1)² (x−3)³
f(x)=−2( x+3 )( x+2 )( x−1 )
f(x)=− 3 2 ( 2x−1 ) 2 ( x−6 )( x+2 )
ਸਥਾਨਕ ਮਹੱਤਮ ( –.58, –.62 ), ਸਥਾਨਕ ਨਿਊਨਤਮ ( .58, –1.38 )
ਸਰਬ-ਪੱਧਰੀ ਨਿਊਨਤਮ ( –.63, –.47 )
ਸਰਬ-ਪੱਧਰੀ ਨਿਊਨਤਮ (.75, .89)
f(x)= (x−500) 2 (x+200)
f(x)=4 x 3 −36 x 2 +80x
f(x)=4 x 3 −36 x 2 +60x+100
f(x)=9π( x 3 +5 x 2 +8x+4)
5.4 ਭਾਗ ਅਭਿਆਸ
ਬਹੁਪਦੀ ਦਾ ਦੋਪਦੀ ਇੱਕ ਗੁਣਨਖੰਡ ਹੈ।
x+6+ 5 x-1 , ਭਾਗਫਲ:x+6, ਬਾਕੀ:5
3x+2, ਭਾਗਫਲ: 3x+2, ਬਾਕੀ: 0
x−5, ਭਾਗਫਲ:x−5, ਬਾਕੀ:0
2x−7+ 16 x+2 , ਭਾਗਫਲ:2x−7, ਬਾਕੀ:16
x−2+ 6 3x+1 , ਭਾਗਫਲ:x−2, ਬਾਕੀ:6
2 x 2 −3x+5, ਭਾਗਫਲ:2 x 2 −3x+5, ਬਾਕੀ:0
2 x 2 +2x+1+ 10 x−4
2 x 2 −7x+1− 2 2x+1
3 x 2 −11x+34− 106 x+3
x 2 +5x+1
4 x 2 −21x+84− 323 x+4
x 2 −14x+49
3 x 2 +x+ 2 3x−1
x³ − 3x + 1
x³ − x² + 2
x³ − 6x² + 12x − 8
x³ − 9x² + 27x − 27
2x³ − 2x + 2
ਹਾਂ (x−2)(3x³ − 5)
ਹਾਂ (x−2)(4x³ + 8x² + x + 2)
ਨਹੀਂ
(x−1)(x² + 2x + 4)
(x−5)(x² + x + 1)
ਭਾਗਫਲ: 4x² + 8x + 16, ਬਾਕੀ: −1
ਭਾਗਫਲ: 3x² + 3x + 5, ਬਾਕੀ: 0
ਭਾਗਫਲ: x³ − 2x² + 4x − 8, ਬਾਕੀ: −6
x⁶ − x⁵ + x⁴ − x³ + x² − x + 1
x³ − x² + x − 1 + 1/(x+1)
1 + 1+i / (x−i)
1 + 1−i / (x+i)
x² + ix − 1 + (1−i) / (x−i)
2x² + 3
2x + 3
x + 2
x − 3
3x² − 2
5.5 ਭਾਗ ਅਭਿਆਸ
ਇਹ ਪ੍ਰਮੇਯ (theorem) ਦੀ ਵਰਤੋਂ ਬਹੁਪਦੀ (polynomial) ਦਾ ਮੁੱਲ ਨਿਰਧਾਰਤ ਕਰਨ ਲਈ ਕੀਤੀ ਜਾ ਸਕਦੀ ਹੈ।
ਪਰਿਮੇਯ ਸਿਫ਼ਰਾਂ (rational zeros) ਨੂੰ ਭਿੰਨਾਂ (fractions) ਵਜੋਂ ਪ੍ਰਗਟ ਕੀਤਾ ਜਾ ਸਕਦਾ ਹੈ, ਜਦੋਂ ਕਿ ਅਸਲ ਸਿਫ਼ਰਾਂ (real zeros) ਵਿੱਚ ਅਪਰਿਮੇਯ ਸੰਖਿਆਵਾਂ (irrational numbers) ਸ਼ਾਮਲ ਹੁੰਦੀਆਂ ਹਨ।
ਬਹੁਪਦੀ ਫੰਕਸ਼ਨਾਂ (polynomial functions) ਵਿੱਚ ਦੁਹਰਾਏ ਗਏ ਸਿਫ਼ਰ (repeated zeros) ਹੋ ਸਕਦੇ ਹਨ, ਇਸ ਲਈ ਇਹ ਤੱਥ ਕਿ ਕੋਈ ਸੰਖਿਆ ਸਿਫ਼ਰ ਹੈ, ਇਸ ਨੂੰ ਦੁਬਾਰਾ ਸਿਫ਼ਰ ਹੋਣ ਤੋਂ ਨਹੀਂ ਰੋਕਦਾ।
−106
0
255
−1
−2, 1, 1 2
−2
−3
− 5 2 , 6 , − 6
2, −4, − 3 2
4, −4, −5
5, −3, − 1 2
1 2 , 1+ 5 2 , 1− 5 2
3 2
2, 3, −1, −2
1 2 , − 1 2 , 2, −3
−1, −1, 5 , − 5
− 3 4 , − 1 2
2, 3+2i, 3−2i
− 2 3 , 1+2i, 1−2i
− 1 2 , 1+4i, 1−4i
1 ਧਨ, 1 ਰਿਣ
3 or 1 positive, 0 negative
0 positive, 3 or 1 negative
2 or 0 positive, 2 or 0 negative
2 or 0 positive, 2 or 0 negative
±5,±1,± 5 2 ,± 1 2 ±5,±1,± 5 2 ,± 1 2
±1, ± 1 2 , ± 1 3 , ± 1 6 ±1, ± 1 2 , ± 1 3 , ± 1 6
1, 1 2 , − 1 3 1, 1 2 , − 1 3
2, 1 4 , − 3 2 2, 1 4 , − 3 2
5 4 5 4
f(x)= 4 9 ( x 3 + x 2 −x−1 ) f(x)= 4 9 ( x 3 + x 2 −x−1 )
f(x)=− 1 5 ( 4 x 3 −x ) f(x)=− 1 5 ( 4 x 3 −x )
8 by 4 by 6 inches
5.5 by 4.5 by 3.5 inches
8 by 5 by 3 inches
Radius = 6 meters, Height = 2 meters
Radius = 2.5 meters, Height = 4.5 meters
5.6 Section Exercises
The rational function will be represented by a quotient of polynomial functions.
The numerator and denominator must have a common factor.
Yes. The numerator of the formula of the functions would have only complex roots and/or factors common to both the numerator and denominator.
All reals x≠–1, 1 All reals x≠–1, 1
All reals x≠–1, –2, 1, 2 All reals x≠–1, –2, 1, 2
V.A. at x=– 2 5 ; x=– 2 5 ; H.A. at y=0; y=0; Domain is all reals x≠– 2 5 x≠– 2 5
V.A. at x=4, –9; x=4, –9; H.A. at y=0; y=0; Domain is all reals x≠4, –9 x≠4, –9
V.A. at x=0, 4, −4; x=0, 4, −4; H.A. at y=0; y=0; Domain is all reals x≠0,4, –4 x≠0,4, –4
V.A. at x=5; x=5; H.A. at y=0; y=0; Domain is all reals x≠5,−5 x≠5,−5
V.A. at x= 1 3 ; x= 1 3 ; H.A. at y=− 2 3 ; y=− 2 3 ; Domain is all reals x≠ 1 3 . x≠ 1 3 .
none
x-intercepts none, y-intercept ( 0, 1 4 ) x-intercepts none, y-intercept ( 0, 1 4 )
Local behavior: x→− 1 2 + ,f(x)→−∞,x→− 1 2 − ,f(x)→∞ x→− 1 2 + ,f(x)→−∞,x→− 1 2 − ,f(x)→∞
End behavior: x→±∞,f(x)→ 1 2 x→±∞,f(x)→ 1 2
Local behavior: x→ 6 + ,f(x)→−∞,x→ 6 − ,f(x)→∞, x→ 6 + ,f(x)→−∞,x→ 6 − ,f(x)→∞, End behavior: x→±∞,f(x)→−2 x→±∞,f(x)→−2
Local behavior: x→ 1 3 + ,f(x)→−∞,x→ 1 3 − , x→ 1 3 + ,f(x)→−∞,x→ 1 3 − , f(x)→∞,x→ - 5 2 + ,f(x)→∞,x→ - 5 2 – ,f(x)→−∞ f(x)→∞,x→ - 5 2 + ,f(x)→∞,x→ - 5 2 – ,f(x)→−∞
End behavior: x→±∞,f(x)→ 1 3 x→±∞,f(x)→ 1 3
y=2x+4 y=2x+4
y=2x y=2x
V.A.x=0,H.A.y=2 V.A.x=0,H.A.y=2
V.A.x=2,H.A.y=0 V.A.x=2,H.A.y=0
V.A.x=−4,H.A.y=2;( 3 2 ,0 );( 0,− 3 4 ) V.A.x=−4,H.A.y=2;( 3 2 ,0 );( 0,− 3 4 )
V.A.x=2,H.A.y=0,(0,1) V.A.x=2,H.A.y=0,(0,1)
V.A.x=−4,x= 4 3 ,H.A.y=1;(5,0);( − 1 3 ,0 );( 0, 5 16 ) V.A.x=−4,x= 4 3 ,H.A.y=1;(5,0);( − 1 3 ,0 );( 0, 5 16 )
V.A.x=−1,H.A.y=1;( −3,0 );( 0,3 ) V.A.x=−1,H.A.y=1;( −3,0 );( 0,3 )
V.A.x=4,S.A.y=2x+9;( −1,0 );( 1 2 ,0 );( 0, 1 4 ) V.A.x=4,S.A.y=2x+9;( −1,0 );( 1 2 ,0 );( 0, 1 4 )
V.A.x=−2,x=4,H.A.y=1,( 1,0 );( 5,0 );( −3,0 );( 0,− 15 16 ) V.A.x=−2,x=4,H.A.y=1,( 1,0 );( 5,0 );( −3,0 );( 0,− 15 16 )
y=50 x 2 −x−2 x 2 −25 y=50 x 2 −x−2 x 2 −25
y=7 x 2 +2x−24 x 2 +9x+20 y=7 x 2 +2x−24 x 2 +9x+20
y= 1 2 x 2 −4x+4 x+1 y= 1 2 x 2 −4x+4 x+1
y=4 x−3 x 2 −x−12 y=4 x−3 x 2 −x−12
y= 27(x−2) (x+3) (x–3)2 y= 27(x−2) (x+3) (x–3)2
y= 1 3 x 2 +x−6 x−1 y= 1 3 x 2 +x−6 x−1
y=−6 (x−1) 2 (x+3) (x−2) 2 y=−6 (x−1) 2 (x+3) (x−2) 2
row: x x | 2.01 | 2.001 | 2.0001 | 1.99 | 1.999
row: y y | 100 | 1,000 | 10,000 | –100 | –1,000
row: x x | 10 | 100 | 1,000 | 10,000 | 100,000
row: y y | .125 | .0102 | .001 | .0001 | .00001
Vertical asymptote x=2, x=2, Horizontal asymptote y=0 y=0
row: x x | –4.1 | –4.01 | –4.001 | –3.99 | –3.999
row: y y | 82 | 802 | 8,002 | –798 | –7998
row: x x | 10 | 100 | 1,000 | 10,000 | 100,000
row: y y | 1.4286 | 1.9331 | 1.992 | 1.9992 | 1.999992
Vertical asymptote x=−4, x=−4, Horizontal asymptote y=2 y=2
row: x x | –.9 | –.99 | –.999 | –1.1 | –1.01
row: y y | 81 | 9,801 | 998,001 | 121 | 10,201
row: x x | 10 | 100 | 1,000 | 10,000 | 100,000
row: y y | .82645 | .9803 | .998 | .9998
Vertical asymptote x=−1, x=−1, Horizontal asymptote y=1 y=1
( 3 2 ,∞ ) ( 3 2 ,∞ )
(−2,1)∪(4,∞) (−2,1)∪(4,∞)
( 2,4 ) ( 2,4 )
( 2,5 ) ( 2,5 )
( –1,1 ) ( –1,1 )
C(t)= 8+2t 300+20t C(t)= 8+2t 300+20t
After about 6.12 hours.
A(x)=50 x 2 + 800 x . A(x)=50 x 2 + 800 x . 2 by 2 by 5 feet.
A(x)=π x 2 + 100 x . A(x)=π x 2 + 100 x . Radius = 2.52 meters.
5.7 Section Exercises
It can be too difficult or impossible to solve for x x in terms of y. y.
We will need a restriction on the domain of the answer.
f −1 (x)= x +4 f −1 (x)= x +4
f −1 (x)= x+3 −1 f −1 (x)= x+3 −1
f −1 (x)= 12−x f −1 (x)= 12−x
f −1 (x)= x−4 2 f −1 (x)= x−4 2
f −1 (x)= x−1 3 3 f −1 (x)= x−1 3 3
f −1 (x)= 4−x 2 3 f −1 (x)= 4−x 2 3
f −1 (x)= 3−x2 4 ,[ 0,∞ ) f −1 (x)= 3−x2 4 ,[ 0,∞ )
f −1 (x)= (x-5)2+8 6 f −1 (x)= (x-5)2+8 6
f −1 (x)= (3-x)2 f −1 (x)=(3-x)2
f −1 (x)= 4x+3 x f −1 (x)= 4x+3 x
f −1 (x)= 7x−3 1−x f −1 (x)= 7x−3 1−x
f −1 (x)= 2x-1 5x+5 f −1 (x)= 2x-1 5x+5
f −1 (x)= x+3 −2 f −1 (x)= x+3 −2
f −1 (x)= x−2 f −1 (x)= x−2
f −1 (x)= x−3 f −1 (x)= x−3
f −1 (x)= x−3 3 f −1 (x)= x−3 3
f −1 (x)= x+4 -2 f −1 (x)= x+4 -2
[-1,0)∪[1,∞) [-1,0)∪[1,∞)
[-3,0]∪(4,∞) [-3,0]∪(4,∞)
[-∞,-4]⋅[-3,3] [-∞,-4]⋅[-3,3]
(–2, 0), (0, 1), (8, 2) (–2, 0), (0, 1), (8, 2)
(–13, –1), (–4, 0), (5, 1) (–13, –1), (–4, 0), (5, 1)
f −1 (x)= x−b a 3 f −1 (x)= x−b a 3
f −1 (x)= x 2 -b a f −1 (x)= x 2 -b a
f −1 (x)= c x - b a - x f −1 (x)= c x - b a - x
t(h)= 600-h 16 t(h)= 600-h 16 , 3.54 seconds
r(A)= A 4π , ≈ r(A)= A 4π , ≈ 8.92 in.
l(T)=32.2(T2π), ≈ l(T)=32.2(T2π), ≈ 3.26 ft
r(A)= A+ 8π 2π r(A)= A+ 8π 2π –2, 3.99 ft
r(V)= V 10π , r(V)= V 10π , ≈ 5.64 ft
5.8 Section Exercises
The graph will have the appearance of a power function.
No. Multiple variables may jointly vary.
y=5 x 2 y=5 x 2
y=11944 x 3 y=11944 x 3
y=6 x 4 y=6 x 4
y= 18 x 2 y= 18 x 2
y= 81 x 4 y= 81 x 4
y= 20 x 3 y= 20 x 3
y=10xzw y=10xzw
y=10x z y=10x z
y=4 xz w y=4 xz w
y=40 xz w t 2 y=40 xz w t 2
y=256 y=256
y=6 y=6
y=6 y=6
y=27 y=27
y=3 y=3
y=18 y=18
y=90 y=90
y= 81 2 y= 81 2
y= 3 4 x 2 y= 3 4 x 2
y= 1 3 x y= 1 3 x
y= 4 x 2 y= 4 x 2
1.89 years
0.61 years
3 seconds
48 inches
49.75 pounds
33.33 amperes
2.88 inches
Review Exercises
f(x)= (x−2) 2 −9vertex (2,–9), intercepts (5,0); (–1,0); (0,–5) f(x)= (x−2) 2 −9vertex (2,–9), intercepts (5,0); (–1,0); (0,–5)
f(x)= 3 25 ( x+2 ) 2 +3 f(x)= 3 25 ( x+2 ) 2 +3
300 meters by 150 meters, the longer side parallel to river.
Yes, degree = 5, leading coefficient = 4
Yes, degree = 4, leading coefficient = 1
Asx→−∞,f(x)→−∞,asx→∞,f(x)→∞ Asx→−∞,f(x)→−∞,asx→∞,f(x)→∞
–3 with multiplicity 2, 1 2 1 2 with multiplicity 1, –1 with multiplicity 3
4 with multiplicity 1
1 2 1 2 with multiplicity 1, 3 with multiplicity 3
x 2 +4 x 2 +4 with remainder 12
x 2 −5x+20− 61 x+3 x 2 −5x+20− 61 x+3
2 x 2 −2x−3 2 x 2 −2x−3 , so factored form is (x+4)(2 x 2 −2x−3) (x+4)(2 x 2 −2x−3)
{ −2, 4, − 1 2 } { −2, 4, − 1 2 }
{ 1, 3, 4, 1 2 } { 1, 3, 4, 1 2 }
0 or 2 positive, 1 negative
Intercepts (–2,0)and( 0,− 2 5 ) (–2,0)and( 0,− 2 5 ) , Asymptotes x=5 x=5 and y=1. y=1.
Intercepts (3, 0), (-3, 0), and ( 0, 27 2 ) ( 0, 27 2 ), Asymptotes x=1, x=–2, y=3. x=1, x=–2, y=3.
y=x−2 y=x−2
f −1 (x)= x +2 f −1 (x)= x +2
f −1 (x)= x+11 −3 f −1 (x)= x+11 −3
f −1 (x)= (x+3) 2 −5 4 ,x≥−3 f −1 (x)= (x+3) 2 −5 4 ,x≥−3
y=64 y=64
y=72 y=72
148.5 pounds
Practice Test
Degree: 5, leading coefficient: −2
As x→−∞, f(x)→∞, As x→∞, f(x)→∞ As x→−∞, f(x)→∞, As x→∞, f(x)→∞
f(x)= 3(x-2)2 f(x)=3(x-2)2
3 with multiplicity 3, 1313 with multiplicity 1, 1 with multiplicity 2
-12-12 with multiplicity 3, 2 with multiplicity 2
x3 + 2x2 + 7x + 14 + 24 x-2 x3+2x2+7x+14+ 24 x-2
{–3,–1,32} {–3,–1,32}
1, −2, and − 3232 (multiplicity 2)
f(x)= -23(x-3)2 (x-1) (x+2) f(x)=-23(x-3)2(x-1)(x+2)
2 or 0 positive, 1 negative
( -3, 0 ) ( 1, 0 ) ( 0, 3 4 ) (-3,0)(1,0)( 0, 3 4 )
f −1 (x)= (x-4)2 +2,x≥4 f −1 (x)= (x-4)2+2,x≥4
f −1 (x)= x+3 3x-2 f −1 (x)= x+3 3x-2
y=20 y=20