5.1 Quadratic Functions
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 − 6 x + 13 g ( x ) = x 2 − 6 x + 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
- ⓒ 7 seconds
5.2 Power Functions and Polynomial Functions
f ( x ) f ( x ) is a power function because it can be written as f ( x ) = 8 x 5 . f ( x ) = 8 x 5 . 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 ) → − ∞ ; a s x → − ∞ , f ( x ) → − ∞ . x → ∞ , f ( x ) → − ∞ ; a s 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.
5.3 Graphs of Polynomial Functions
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 ) .
5.4 Dividing Polynomials
4 x 2 − 8 x + 15 − 78 4 x + 5 4 x 2 − 8 x + 15 − 78 4 x + 5
3 x 3 − 3 x 2 + 21 x − 150 + 1 , 090 x + 7 3 x 3 − 3 x 2 + 21 x − 150 + 1 , 090 x + 7
3 x 2 − 4 x + 1 3 x 2 − 4 x + 1
5.5 Zeros of Polynomial Functions
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 − 2 x + 10 f ( x ) = − 1 2 x 3 + 5 2 x 2 − 2 x + 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
5.6 Rational Functions
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 − 6 x + 9 ) ( x − 3 ) ( x − 3 ) = − 4 x 2 + 24 x − 35 x 2 − 6 x + 9 f ( x ) = 1 ( x − 3 ) 2 − 4 = 1 − 4 ( x − 3 ) 2 ( x − 3 ) 2 = 1 − 4 ( x 2 − 6 x + 9 ) ( x − 3 ) ( x − 3 ) = − 4 x 2 + 24 x − 35 x 2 − 6 x + 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.
5.7 Inverses and Radical Functions
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 ( 3 x − 5 ) = ( 3 x − 5 ) + 5 3 = 3 x 3 = x f ( f − 1 ( x ) ) = f ( 3 x − 5 ) = ( 3 x − 5 ) + 5 3 = 3 x 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 ) = 2 x + 3 x − 1 f − 1 ( x ) = 2 x + 3 x − 1
5.8 Modeling Using Variation
128 3 128 3
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 + 4 x + 3 f ( x ) = x 2 + 4 x + 3
f ( x ) = x 2 - 4 x + 7 f ( x ) = x 2 - 4 x + 7
f ( x ) = - 1 49 x 2 + 6 49 x + 89 49 f ( x ) = - 1 49 x 2 + 6 49 x + 89 49
f ( x ) = x 2 - 2 x + 1 f ( x ) = x 2 - 2 x + 1
Vertex: (3, −10), axis of symmetry: x = 3, intercepts: ( 3 + 10 , 0 ) ( 3 + 10 , 0 ) and ( 3 - 10 , 0 ) ( 3 - 10 , 0 )
Vertex: ( 3 2 , - 12 ) ( 3 2 , - 12 ) , axis of symmetry: x = 3 2 x = 3 2 , intercept: ( 3 + 2 3 2 , 0 ) ( 3 + 2 3 2 , 0 ) and ( 3 - 2 3 2 , 0 ) ( 3 - 2 3 2 , 0 )
f ( x ) = x 2 + 2 x + 3 f ( x ) = x 2 + 2 x + 3
f ( x ) = - 3 x 2 − 6 x − 1 f ( x ) = - 3 x 2 − 6 x − 1
f ( x ) = - 1 4 x 2 − x + 2 f ( x ) = - 1 4 x 2 − x + 2
f ( x ) = x 2 + 2 x + 1 f ( x ) = x 2 + 2 x + 1
f ( x ) = - x 2 + 2 x f ( x ) = - x 2 + 2 x
f ( x ) = 2 x 2 f ( x ) = 2 x 2
The graph is shifted to the right or left (a horizontal shift).
The suspension bridge has 1,000 feet distance from the center.
Domain is ( − ∞ , ∞ ) . ( − ∞ , ∞ ) . Range is ( - ∞ , 2 ] . ( - ∞ , 2 ] .
Domain: ( - ∞ , ∞ ) ( - ∞ , ∞ ) ; range: [ 100 , ∞ ) [ 100 , ∞ )
f ( x ) = 2 x 2 + 2 f ( x ) = 2 x 2 + 2
f ( x ) = - x 2 − 2 f ( x ) = - x 2 − 2
f ( x ) = 3 x 2 + 6 x - 15 f ( x ) = 3 x 2 + 6 x - 15
75 feet by 50 feet
3 and 3; product is 9
The revenue reaches the maximum value when 1800 thousand phones are produced.
2.449 seconds
41 trees per acre
5.2 Section Exercises
The coefficient of the power function is the real number that is multiplied by the variable raised to a power. The degree is the highest power appearing in the function.
As x x decreases without bound, so does f ( x ) . f ( x ) . As x x increases without bound, so does f ( x ) . f ( x ) .
The polynomial function is of even degree and leading coefficient is negative.
Power function
Degree = 2, Coefficient = –2
Degree =4, Coefficient = –2
As x → ∞ x → ∞ , f ( x ) → ∞ , as x → − ∞ , f ( x ) → ∞ f ( x ) → ∞ , as x → − ∞ , f ( x ) → ∞
As x → − ∞ x → − ∞ , f ( x ) → − ∞ , as x → ∞ , f ( x ) → − ∞ f ( x ) → − ∞ , as x → ∞ , f ( x ) → − ∞
As x → ∞ x → ∞ , f ( x ) → ∞ , as x → − ∞ , f ( x ) → − ∞ f ( x ) → ∞ , as x → − ∞ , f ( x ) → − ∞
y -intercept is ( 0 , 12 ) , ( 0 , 12 ) , t -intercepts are ( 1 , 0 ) ; ( – 2 , 0 ) ; and ( 3 , 0 ) . ( 1 , 0 ) ; ( – 2 , 0 ) ; and ( 3 , 0 ) .
y -intercept is ( 0 , − 16 ) . ( 0 , − 16 ) . x -intercepts are ( 2 , 0 ) ( 2 , 0 ) and ( − 2 , 0 ) . ( − 2 , 0 ) .
y -intercept is ( 0 , 0 ) . ( 0 , 0 ) . x -intercepts are ( 0 , 0 ) , ( 4 , 0 ) , ( 0 , 0 ) , ( 4 , 0 ) , and ( − 2 , 0 ) . ( − 2 , 0 ) .
Yes. Number of turning points is 2. Least possible degree is 3.
Yes. Number of turning points is 1. Least possible degree is 2.
Yes. Number of turning points is 0. Least possible degree is 1.
As x → − ∞ x → − ∞ , f ( x ) → ∞ , as x → ∞ , f ( x ) → ∞ f ( x ) → ∞ , as x → ∞ , f ( x ) → ∞
As x → − ∞ x → − ∞ , f ( x ) → ∞ , as x → ∞ , f ( x ) → − ∞ f ( x ) → ∞ , as x → ∞ , 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 ) → ∞ , as x → ∞ , f ( x ) → ∞ f ( x ) → ∞ , as x → ∞ , 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 ) → − ∞ , as x → ∞ , f ( x ) → ∞ f ( x ) → − ∞ , as x → ∞ , 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 ) . A s x → − ∞ A s x → − ∞ , f ( x ) → − ∞ , as x → ∞ , f ( x ) → ∞ f ( x ) → − ∞ , as x → ∞ , 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 ) → ∞ , as x → ∞ , f ( x ) → ∞ f ( x ) → ∞ , as x → ∞ , 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 ) → − ∞ , as x → ∞ , f ( x ) → ∞ f ( x ) → − ∞ , as x → ∞ , f ( x ) → ∞
f ( x ) = x 2 − 4 f ( x ) = x 2 − 4
f ( x ) = x 3 − 4 x 2 + 4 x f ( x ) = x 3 − 4 x 2 + 4 x
f ( x ) = x 4 + 1 f ( x ) = x 4 + 1
V ( m ) = 8 m 3 + 36 m 2 + 54 m + 27 V ( m ) = 8 m 3 + 36 m 2 + 54 m + 27
V ( x ) = 4 x 3 − 32 x 2 + 64 x V ( x ) = 4 x 3 − 32 x 2 + 64 x
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 and f ( 4 ) = 28. f ( 4 ) = 28. Sign change confirms.
f ( 1 ) = 3 f ( 1 ) = 3 and f ( 3 ) = – 77. f ( 3 ) = – 77. Sign change confirms.
f ( 0.01 ) = 1.000001 f ( 0.01 ) = 1.000001 and f ( 0.1 ) = – 7.999. f ( 0.1 ) = – 7.999. Sign change confirms.
0 with multiplicity 2, − 3 2 − 3 2 with multiplicity 5, 4 with multiplicity 2
0 with multiplicity 2, –2 with multiplicity 2
− 2 3 − 2 3 with multiplicity 5, 5 with multiplicity 2
0 with multiplicity 4, 2 with multiplicity 1, −1 with multiplicity 1
3 2 3 2 with multiplicity 2, 0 with multiplicity 3
0 with multiplicity 6 , 2 3 with multiplicity 2 0 with multiplicity 6 , 2 3 with multiplicity 2
x -intercepts, ( 1, 0 ) ( 1, 0 ) with multiplicity 2, ( – 4 , 0 ) ( – 4 , 0 ) with multiplicity 1, y - y - intercept ( 0 , 4 ). ( 0 , 4 ). As x → − ∞ , f ( x ) → − ∞ , as x → ∞ , f ( x ) → ∞ . x → − ∞ , f ( x ) → − ∞ , as x → ∞ , f ( x ) → ∞ .
x -intercepts ( 3 , 0 ) ( 3 , 0 ) with multiplicity 3, ( 2 , 0 ) ( 2 , 0 ) with multiplicity 2, y - y - intercept ( 0 , – 108 ) . ( 0 , – 108 ) . As x → − ∞ , f ( x ) → − ∞ , as x → ∞ , f ( x ) → ∞ . x → − ∞ , f ( x ) → − ∞ , as x → ∞ , f ( x ) → ∞ .
x -intercepts ( 0 , 0 ) , ( – 2 , 0 ) , ( 4 , 0 ) ( 0 , 0 ) , ( – 2 , 0 ) , ( 4 , 0 ) with multiplicity 1, y - y - intercept ( 0 , 0 ) . ( 0 , 0 ) . As x → − ∞ , f ( x ) → ∞ , as x → ∞ , f ( x ) → − ∞ . x → − ∞ , f ( x ) → ∞ , as x → ∞ , f ( x ) → − ∞ .
f ( x ) = − 2 9 ( x − 3 ) ( x + 1 ) ( x + 3 ) f ( x ) = − 2 9 ( x − 3 ) ( x + 1 ) ( x + 3 )
f ( x ) = 1 4 ( x + 2 ) 2 ( x − 3 ) f ( x ) = 1 4 ( x + 2 ) 2 ( x − 3 )
–4, –2, 1, 3 with multiplicity 1
–2, 3 each with multiplicity 2
f ( x ) = − 2 3 ( x + 2 ) ( x − 1 ) ( x − 3 ) f ( x ) = − 2 3 ( x + 2 ) ( x − 1 ) ( x − 3 )
f ( x ) = 1 3 ( x − 3 ) 2 ( x − 1 ) 2 ( 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 ) = −15 ( x − 1 ) 2 ( x − 3 ) 3
f ( x ) = − 2 ( x + 3 ) ( x + 2 ) ( x − 1 ) f ( x ) = − 2 ( x + 3 ) ( x + 2 ) ( x − 1 )
f ( x ) = − 3 2 ( 2 x − 1 ) 2 ( x − 6 ) ( x + 2 ) f ( x ) = − 3 2 ( 2 x − 1 ) 2 ( x − 6 ) ( x + 2 )
local max ( – .58, – .62 ) , ( – .58, – .62 ) , local min ( .58, –1 .38 ) ( .58, –1 .38 )
global min ( – .63, – .47 ) ( – .63, – .47 )
global min ( .75, .89) ( .75, .89)
f ( x ) = ( x − 500 ) 2 ( x + 200 ) f ( x ) = ( x − 500 ) 2 ( x + 200 )
f ( x ) = 4 x 3 − 36 x 2 + 80 x f ( x ) = 4 x 3 − 36 x 2 + 80 x
f ( x ) = 4 x 3 − 36 x 2 + 60 x + 100 f ( x ) = 4 x 3 − 36 x 2 + 60 x + 100
f ( x ) = 9 π ( x 3 + 5 x 2 + 8 x + 4 ) f ( x ) = 9 π ( x 3 + 5 x 2 + 8 x + 4 )
5.4 Section Exercises
The binomial is a factor of the polynomial.
x + 6 + 5 x - 1 , quotient: x + 6 , remainder: 5 x + 6 + 5 x - 1 , quotient: x + 6 , remainder: 5
3 x + 2 , quotient: 3 x + 2 , remainder: 0 3 x + 2 , quotient: 3 x + 2 , remainder: 0
x − 5 , quotient: x − 5 , remainder: 0 x − 5 , quotient: x − 5 , remainder: 0
2 x − 7 + 16 x + 2 , quotient: 2 x − 7 , remainder: 16 2 x − 7 + 16 x + 2 , quotient: 2 x − 7 , remainder: 16
x − 2 + 6 3 x + 1 , quotient: x − 2 , remainder: 6 x − 2 + 6 3 x + 1 , quotient: x − 2 , remainder: 6
2 x 2 − 3 x + 5 , quotient: 2 x 2 − 3 x + 5 , remainder: 0 2 x 2 − 3 x + 5 , quotient: 2 x 2 − 3 x + 5 , remainder: 0
2 x 2 + 2 x + 1 + 10 x − 4 2 x 2 + 2 x + 1 + 10 x − 4
2 x 2 − 7 x + 1 − 2 2 x + 1 2 x 2 − 7 x + 1 − 2 2 x + 1
3 x 2 − 11 x + 34 − 106 x + 3 3 x 2 − 11 x + 34 − 106 x + 3
x 2 + 5 x + 1 x 2 + 5 x + 1
4 x 2 − 21 x + 84 − 323 x + 4 4 x 2 − 21 x + 84 − 323 x + 4
x 2 − 14 x + 49 x 2 − 14 x + 49
3 x 2 + x + 2 3 x − 1 3 x 2 + x + 2 3 x − 1
x 3 − 3 x + 1 x 3 − 3 x + 1
x 3 − x 2 + 2 x 3 − x 2 + 2
x 3 − 6 x 2 + 12 x − 8 x 3 − 6 x 2 + 12 x − 8
x 3 − 9 x 2 + 27 x − 27 x 3 − 9 x 2 + 27 x − 27
2 x 3 − 2 x + 2 2 x 3 − 2 x + 2
Yes ( x − 2 ) ( 3 x 3 − 5 ) ( x − 2 ) ( 3 x 3 − 5 )
Yes ( x − 2 ) ( 4 x 3 + 8 x 2 + x + 2 ) ( x − 2 ) ( 4 x 3 + 8 x 2 + x + 2 )
( x − 1 ) ( x 2 + 2 x + 4 ) ( x − 1 ) ( x 2 + 2 x + 4 )
( x − 5 ) ( x 2 + x + 1 ) ( x − 5 ) ( x 2 + x + 1 )
Quotient: 4 x 2 + 8 x + 16 , remainder: − 1 Quotient: 4 x 2 + 8 x + 16 , remainder: − 1
Quotient: 3 x 2 + 3 x + 5 , remainder: 0 Quotient: 3 x 2 + 3 x + 5 , remainder: 0
Quotient: x 3 − 2 x 2 + 4 x − 8 , remainder: − 6 Quotient: x 3 − 2 x 2 + 4 x − 8 , remainder: − 6
x 6 − x 5 + x 4 − x 3 + x 2 − x + 1 x 6 − x 5 + x 4 − x 3 + x 2 − x + 1
x 3 − x 2 + x − 1 + 1 x + 1 x 3 − x 2 + x − 1 + 1 x + 1
1 + 1 + i x − i 1 + 1 + i x − i
1 + 1 − i x + i 1 + 1 − i x + i
x 2 − i x − 1 + 1 − i x − i x 2 − i x − 1 + 1 − i x − i
2 x 2 + 3 2 x 2 + 3
2 x + 3 2 x + 3
x + 2 x + 2
x − 3 x − 3
3 x 2 − 2 3 x 2 − 2
5.5 Section Exercises
The theorem can be used to evaluate a polynomial.
Rational zeros can be expressed as fractions whereas real zeros include irrational numbers.
Polynomial functions can have repeated zeros, so the fact that number is a zero doesn’t preclude it being a zero again.
− 106 − 106
− 2 , 1 , 1 2 − 2 , 1 , 1 2
− 5 2 , 6 , − 6 − 5 2 , 6 , − 6
2 , − 4 , − 3 2 2 , − 4 , − 3 2
4 , − 4 , − 5 4 , − 4 , − 5
5 , − 3 , − 1 2 5 , − 3 , − 1 2
1 2 , 1 + 5 2 , 1 − 5 2 1 2 , 1 + 5 2 , 1 − 5 2
2 , 3 , − 1 , − 2 2 , 3 , − 1 , − 2
1 2 , − 1 2 , 2 , − 3 1 2 , − 1 2 , 2 , − 3
− 1 , − 1 , 5 , − 5 − 1 , − 1 , 5 , − 5
− 3 4 , − 1 2 − 3 4 , − 1 2
2 , 3 + 2 i , 3 − 2 i 2 , 3 + 2 i , 3 − 2 i
− 2 3 , 1 + 2 i , 1 − 2 i − 2 3 , 1 + 2 i , 1 − 2 i
− 1 2 , 1 + 4 i , 1 − 4 i − 1 2 , 1 + 4 i , 1 − 4 i
1 positive, 1 negative
3 or 1 positive, 0 negative
0 positive, 3 or 1 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
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 .
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 = 2 x + 4 y = 2 x + 4
y = 2 x y = 2 x
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 = 2 x + 9 ; ( − 1 , 0 ) ; ( 1 2 , 0 ) ; ( 0 , 1 4 ) V . A . x = 4 , S . A . y = 2 x + 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 + 2 x − 24 x 2 + 9 x + 20 y = 7 x 2 + 2 x − 24 x 2 + 9 x + 20
y = 1 2 x 2 − 4 x + 4 x + 1 y = 1 2 x 2 − 4 x + 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
Vertical asymptote x = 2 , x = 2 , Horizontal asymptote y = 0 y = 0
Vertical asymptote x = − 4 , x = − 4 , Horizontal asymptote y = 2 y = 2
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 + 2 t 300 + 20 t C ( t ) = 8 + 2 t 300 + 20 t
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 − x 2 4 , [ 0 , ∞ ) f −1 ( x ) = 3 − x 2 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 ) = 4 x + 3 x f −1 ( x ) = 4 x + 3 x
f −1 ( x ) = 7 x − 3 1 − x f −1 ( x ) = 7 x − 3 1 − x
f −1 ( x ) = 2 x - 1 5 x + 5 f −1 ( x ) = 2 x - 1 5 x + 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 4 π , ≈ r ( A ) = A 4 π , ≈ 8.92 in.
l ( T ) = 32.2 ( T 2 π ) , ≈ l ( T ) = 32.2 ( T 2 π ) , ≈ 3.26 ft
r ( A ) = A + 2 π 8 π , r ( A ) = A + 2 π 8 π , -2, 3.99 ft
r ( V ) = V 10 π , r ( V ) = V 10 π , ≈ 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 = 1 1944 x 3 y = 1 1944 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 = 10 x z w y = 10 x z w
y = 10 x z y = 10 x z
y = 4 x z w y = 4 x z w
y = 40 x z w t 2 y = 40 x z w t 2
y = 256 y = 256
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
49.75 pounds
33.33 amperes
2.88 inches
Review Exercises
f ( x ) = ( x − 2 ) 2 − 9 vertex ( 2 , –9 ) , intercepts ( 5 , 0 ) ; ( –1 , 0 ) ; ( 0 , –5 ) f ( x ) = ( x − 2 ) 2 − 9 vertex ( 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
As x → − ∞ , f ( x ) → − ∞ , as x → ∞ , f ( x ) → ∞ As x → − ∞ , f ( x ) → − ∞ , as x → ∞ , 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 − 5 x + 20 − 61 x + 3 x 2 − 5 x + 20 − 61 x + 3
2 x 2 − 2 x − 3 2 x 2 − 2 x − 3 , so factored form is ( x + 4 ) ( 2 x 2 − 2 x − 3 ) ( x + 4 ) ( 2 x 2 − 2 x − 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, 1 3 1 3 with multiplicity 1, 1 with multiplicity 2
- 1 2 - 1 2 with multiplicity 3, 2 with multiplicity 2
x 3 + 2 x 2 + 7 x + 14 + 26 x - 2 x 3 + 2 x 2 + 7 x + 14 + 26 x - 2
{ –3 , –1 , 3 2 } { –3 , –1 , 3 2 }
1, −2, and − 3 2 3 2 (multiplicity 2)
f ( x ) = - 2 3 ( x - 3 ) 2 ( x - 1 ) ( x + 2 ) f ( x ) = - 2 3 ( 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 3 x - 2 f − 1 ( x ) = x + 3 3 x - 2
y = 20 y = 20
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