Question 1

Determine the maximum number of turning points of f. -f(x) = -x² (x + 4)3(x² - 1) A) 6 B) 2 C) 5 D) 7

Accepted Answer

f(x) has a maximum of 6 turning points because it is a polynomial of degree 6.

Question 2

Find the x- and y-intercepts of f.
-$$f ( x ) = 6 x - x ^ { 3 }$$
A) x-intercepts: $0 , \sqrt { 6 } , - \sqrt { 6 } ; y$-intercept: 0
B) x-intercepts: $0 , - 6 ; y$-intercept: 0
C) x-intercepts: $0 , - 6 ; y$-intercept: 6
D) $x$-intercepts: $0 , \sqrt { 6 } , - \sqrt { 6 } ; y$-intercept: 6

Accepted Answer

To find the x-intercepts, set $f(x) = 0$: $0 = 6x - x^3$. Factoring out $x$, we get $x(6 - x^2) = 0$, which gives $x = 0$, $x = \sqrt{6}$, and $x = -\sqrt{6}$. For the y-intercept, set $x = 0$: $f(0) = 6(0) - (0)^3 = 0$.

Question 3

$$f ( x ) = 5 \left( x ^ { 2 } + 6 \right) \left( x ^ { 2 } + 4 \right) ^ { 2 }$$
A) $\sqrt { 6 }$, multiplicity 1 , crosses $x$-axis; $- \sqrt { 6 }$, multiplicity 1 , crosses $x$-axis; 2 , multiplicity 2 , touches $x$-axis; $- 2$, multiplicity 2 , touches $x$-axis
B) -6, multiplicity 1 , touches $x$-axis; $- 4$, multiplicity 2, crosses $x$-axis
C) -6, multiplicity 1 , crosses $x$-axis; $- 4$, multiplicity 2 , touches $x$-axis
D) No real zeros

Accepted Answer

The function $f(x)$ is a product of two quadratic expressions, each of which has no real roots. Therefore, $f(x)$ has no real zeros.

Question 4

Determine the maximum number of turning points of f.
-$$f ( x ) = \left( x + \frac { 1 } { 9 } \right) ^ { 4 } ( x - 4 ) ^ { 3 }$$
A) above the $x$-axis: $( 4 , \infty )$
below the $x$-axis: $\left( - \infty , - \frac { 1 } { 9 } \right) , \left( - \frac { 1 } { 9 } , 4 \right)$
B) above the $x$-axis: $\left( - \infty , - \frac { 1 } { 9 } \right) , \left( - \frac { 1 } { 9 } , 4 \right)$
below the $x$-axis: $( 4 , \infty )$
C) above the $x$-axis: $\left( - \frac { 1 } { 9 } , 4 \right)$
below the $x$-axis: $\left( - \infty , - \frac { 1 } { 9 } \right) , ( 4 , \infty )$
D) above the x-axis: $\left( - \infty , - \frac { 1 } { 9 } \right) , ( 4 , \infty )$
below the $x$-axis: $\left( - \frac { 1 } { 9 } , 4 \right)$

Accepted Answer

The maximum number of turning points of a polynomial function is equal to the degree of the polynomial minus 1. The degree of the given polynomial is 7, so the maximum number of turning points is 6. The polynomial is positive for $x > 4$ and negative for $x < - \frac { 1 } { 9 }$, so it has 3 turning points above the $x$-axis and 3 turning points below the $x$-axis.

Question 5

Find the x- and y-intercepts of f.
-$$f ( x ) = - x ^ { 2 } ( x + 6 ) \left( x ^ { 2 } - 1 \right)$$
A) x-intercepts: -6, 0, 1; y-intercept: -6
B) x-intercepts: -1, 0, 1, 6; y-intercept: 0
C) x-intercepts: -6, -1, 0, 1; y-intercept: 0
D) x-intercepts: -6, -1, 0, 1; y-intercept: -6

Accepted Answer

To find the x-intercepts, set $f(x) = 0$ and solve for $x$. The given function is $f(x) = -x^2(x + 6)(x^2 - 1)$. Setting this equal to zero, we find the roots are $x = 0$, $x = -6$, and $x^2 - 1 = 0$ which gives $x = 1$ and $x = -1$. Therefore, the x-intercepts are -6, -1, 0, and 1. To find the y-intercept, set $x = 0$ and solve for $f(x)$, which gives $f(0) = 0$. Hence, the y-intercept is 0.

Question 6

Find the x- and y-intercepts of f.
-f(x) = (x - 3)(x - 1)
A) x-intercepts: 3, 1; y-intercept: -4
B) x-intercepts: -3, -1; y-intercept: -4
C) x-intercepts: 3, 1; y-intercept: 3
D) x-intercepts: -3, -1; y-intercept: 3

Accepted Answer

The answer of Find the x- and y-intercepts of f.
-f(x)...

Question 7

Determine the maximum number of turning points of f. -f(x) = 8x - x³ A) 1 B) 3 C) 2 D) 4

Accepted Answer

The given function is $f(x) = 8x - x^{13}$. The maximum number of turning points of a polynomial function is one less than its degree. Since the highest degree of the polynomial is 13, the maximum number of turning points is $13 - 1 = 12$. However, none of the provided options correctly matches this calculation. Given the options, the closest correct procedure would involve recognizing the error in the options relative to the standard mathematical principle.

Question 8

$$f ( x ) = \frac { 1 } { 3 } x ^ { 2 } \left( x ^ { 2 } - 5 \right) ( x - 3 )$$
A) 0 , multiplicity 2 , touches $x$-axis;
3 , multiplicity 1 , crosses $x$-axis;
$\sqrt { 5 }$, multiplicity 1 , crosses $x$-axis
$- \sqrt { 5 }$, multiplicity 1 , crosses $x$-axis 
B) 0 , multiplicity 2 , crosses $x$-axis;
 3 , multiplicity 1 , touches $x$-axis;
$\sqrt { 5 }$, multiplicity 1 , touches $x$-axis; 
$- \sqrt { 5 }$, multiplicity 1 , touches $x$-axis
C) 0 , multiplicity 2, crosses x-axis;
 3 , multiplicity 1 , touches $x$-axis
D) 0 , multiplicity 2 , touches $x$-axis; 
3 , multiplicity 1 , crosses $x$-axis

Accepted Answer

The function $f(x) = \frac{1}{3}x^2(x^2 - 5)(x - 3)$ has zeros at $x = 0$, $x = 3$, $x = \sqrt{5}$, and $x = -\sqrt{5}$. The zero at $x = 0$ has multiplicity 2 because $x^2$ is squared, indicating the graph touches the x-axis at this point. The zeros at $x = 3$, $x = \sqrt{5}$, and $x = -\sqrt{5}$ each have multiplicity 1, meaning the graph crosses the x-axis at these points.

Question 9

$$f ( x ) = 4 - ( x + 4 ) ^ { 5 }$$

A)

B)

C)

D)

Accepted Answer

The answer of \[f ( x ) = 4 -...

Question 10

$f ( x ) = - 2 ( x - 5 ) ^ { 5 } + 3$

A)

B)

C)

D)

Accepted Answer

The answer of \(f ( x ) = - 2...

Question 11

Find the x- and y-intercepts of f.
-$$f ( x ) = 2 x ^ { 4 } ( x - 5 ) ^ { 5 }$$
A) x-intercepts: 0, -5; y-intercept: 2
B) x-intercepts: 0, 5; y-intercept: 2
C) x-intercepts: 0, -5; y-intercept: 0
D) x-intercepts: 0, 5; y-intercept: 0

Accepted Answer

To find the x-intercepts, set $f(x) = 0$. This gives $2x^4(x - 5)^5 = 0$, which has solutions at $x = 0$ and $x = 5$. For the y-intercept, set $x = 0$ in $f(x)$, which gives $f(0) = 2(0)^4(0 - 5)^5 = 0$.

Question 12

Determine the maximum number of turning points of f. -f(x) = (x - 3)²(x + 5)² A) 3 B) 1 C) 4 D) 2

Accepted Answer

The degree of the polynomial is 4 (since (x-3)^2(x+5)^2 expands to a polynomial of degree 4), and the maximum number of turning points for a polynomial is one less than its degree, which is 3.

Question 13

Find the x- and y-intercepts of f.
-$$f ( x ) = - x ^ { 2 } ( x + 6 ) \left( x ^ { 2 } + 1 \right)$$ $$( x ) = - x ^ { 2 } ( x + 6 ) \left( x ^ { 2 } + 1 \right)$$
A) x-intercepts: -6, -1, 0; y-intercept: 6
B) x-intercepts: -6, -1, 0; y-intercept: -6
C) x-intercepts: -6, -1, 0, 1; y-intercept: 0
D) x-intercepts: -6, 0; y-intercept: 0

Accepted Answer

To find the x-intercepts, set $f(x) = 0$. The equation becomes $-x^2(x + 6)(x^2 + 1) = 0$. The solutions are $x = 0$ and $x = -6$. The term $x^2 + 1 = 0$ has no real solutions because $x^2 + 1 > 0$ for all real x. To find the y-intercept, set $x = 0$ in $f(x)$, which gives $f(0) = 0$. Therefore, the x-intercepts are -6 and 0, and the y-intercept is 0.

Question 14

Determine the maximum number of turning points of f.
-$$f ( x ) = x ^ { 2 } ( x + 2 )$$
Answer: (a) For large values of $| x |$, the graph of $f ( x )$ will resemble the graph of $y = x ^ { 3 }$.
(b) $y$-intercept: $( 0,0 )$, $x$-intercepts: $( 0,0 )$ and $( - 2,0 )$
(c) The graph of $f$ crosses the $x$-axis at $( - 2,0 )$ and touches the $x$-axis at $( 0,0 )$.
(e) Local minimum at $( 0,0 )$, Local maximum at $( - 1.33,1.19 )$
(f)

(g) Domain of f: all real numbers; range of f: all real numbers
(h) f is increasing on (-, -1.33) and (0, ); f is decreasing on (-1.33, 0)

Accepted Answer

The answer of Determine the maximum number of turning points...

Question 15

$$f ( x ) = \frac { 1 } { 5 } x ^ { 4 } \left( x ^ { 2 } - 5 \right)$$
A) 0 , multiplicity 4 , touches $x$-axis;
$\sqrt { 5 }$, multiplicity 1 , crosses $x$-axis;
$- \sqrt { 5 }$, multiplicity 1 , crosses $x$-axis
B) 0 , multiplicity 4 , crosses $x$-axis
C) 0 , multiplicity 4 , crosses $x$-axis;
$\sqrt { 5 }$, multiplicity 1 , touches $x$-axis;
D) 0 , multiplicity 4 , touches $x$-axis
$- \sqrt { 5 }$, multiplicity 1 , touches $x$-axis

Accepted Answer

The function $f(x) = \frac{1}{5}x^4(x^2 - 5)$ has roots at $x = 0$, $x = \sqrt{5}$, and $x = -\sqrt{5}$. The root at $x = 0$ has multiplicity 4 because $x^4$ is a factor, indicating the graph touches the x-axis at this point. The roots at $x = \sqrt{5}$ and $x = -\sqrt{5}$ each have multiplicity 1, indicating the graph crosses the x-axis at these points.

Question 16

Determine the maximum number of turning points of f.
-$$f ( x ) = ( x - 4 ) ^ { 3 }$$
A) above the $x$-axis: $( 4 , \infty )$
below the $x$-axis: $( - \infty , 4 )$
B) above the $x$-axis: $( - \infty , 4 ) , ( 4 , \infty )$
below the $x$-axis: no intervals
C) above the x-axis: no intervals
below the $x$-axis: $( - \infty , 4 ) , ( 4 , \infty )$
D) above the $x$-axis: $( - \infty , 4 )$
below the $x$-axis: $( 4 , \infty )$

Accepted Answer

The function $f(x) = (x - 4)^3$ is a cubic function with one real root at $x = 4$. Since it is an odd-degree polynomial, it will extend from the lower left quadrant to the upper right quadrant of the Cartesian plane, meaning it will be below the $x$-axis when $x < 4$ and above the $x$-axis when $x > 4$. There are no turning points because the graph is monotonic, continuously increasing or decreasing without any bends.

Question 17

Find the x- and y-intercepts of f.
-$$f ( x ) = ( x + 5 ) ( x - 3 ) ( x + 3 )$$
A) x-intercepts: -3, 3, 5; y-intercept: -45
B) x-intercepts: -5, -3, 3; y-intercept: -45
C) x-intercepts: -5, -3, 3; y-intercept: 45
D) x-intercepts: -3, 3, 5; y-intercept: 45

Accepted Answer

The x-intercepts are found by setting $f(x) = 0$, which gives the solutions $x = -5$, $x = 3$, and $x = -3$. The y-intercept is found by setting $x = 0$, which gives $f(0) = (0 + 5)(0 - 3)(0 + 3) = 5 \times -3 \times 3 = -45$.

Question 18

Find the x- and y-intercepts of f.
-$$f ( x ) = ( x + 1 ) ( x - 2 ) ( x - 1 ) ^ { 2 }$$
A) x-intercepts: -1, 1, -2; y-intercept: 2
B) x-intercepts: -1, 1, 2; y-intercept: 2
C) x-intercepts: -1, 1, -2; y-intercept: -2
D) x-intercepts: -1, 1, 2; y-intercept: -2

Accepted Answer

The x-intercepts are found by setting $f(x) = 0$, which gives the roots $x = -1$, $x = 2$, and $x = 1$. The y-intercept is found by setting $x = 0$ in $f(x)$, which gives $f(0) = (0 + 1)(0 - 2)(0 - 1)^2 = -2$.

Question 19

Find the x- and y-intercepts of f.
-$$f ( x ) = ( x + 6 ) ^ { 2 }$$
A) x-intercept: -6; y-intercept: 36
B) x-intercept: -6; y-intercept: 0
C) x-intercept: 6; y-intercept: 36
D) x-intercept: 6; y-intercept: 0

Accepted Answer

The x-intercept is found by setting $f(x) = 0$, which gives $x = -6$ after solving $(x + 6)^2 = 0$. The y-intercept is found by setting $x = 0$, which gives $f(0) = (0 + 6)^2 = 36$.

Question 20

Determine the maximum number of turning points of f.
-$$f ( x ) = ( x - 3 ) ^ { 2 } ( x + 5 ) ^ { 2 }$$
A) above the $x$-axis: $( - \infty , - 5 ) , ( 3 , \infty )$
below the x-axis: $( - 5,3 )$
B) above the x-axis: $( - 5,3 )$
below the $x$-axis: $( - \infty , - 5 ) , ( 3 , \infty )$
C) above the $x$-axis: $( - \infty , - 5 ) , ( - 5,3 ) , ( 3 , \infty )$
below the $x$-axis: no intervals
D) above the $x$-axis: no intervals
below the $x$-axis: $( - \infty , - 5 ) , ( - 5,3 ) , ( 3 , \infty )$

Accepted Answer

The function $f(x) = (x - 3)^2(x + 5)^2$ is always non-negative because it is the square of two terms, meaning it cannot be below the x-axis. The function touches the x-axis at $x = 3$ and $x = -5$, but does not cross it, indicating that all intervals are above the x-axis.

Quiz 3: Polynomial and Rational Functions

Determine the maximum number of turning points of f. -f(x) = -x² (x + 4) 3(x² - 1)

Find the x- and y-intercepts of f. - $f ( x ) = 6 x - x ^ { 3 }$

$f ( x ) = 5 \left( x ^ { 2 } + 6 \right) \left( x ^ { 2 } + 4 \right) ^ { 2 }$

Determine the maximum number of turning points of f. - $f ( x ) = \left( x + \frac { 1 } { 9 } \right) ^ { 4 } ( x - 4 ) ^ { 3 }$

Find the x- and y-intercepts of f. - $f ( x ) = - x ^ { 2 } ( x + 6 ) \left( x ^ { 2 } - 1 \right)$

Find the x- and y-intercepts of f. -f(x) = (x - 3) (x - 1)

Determine the maximum number of turning points of f. -f(x) = 8x - x³

$f ( x ) = \frac { 1 } { 3 } x ^ { 2 } \left( x ^ { 2 } - 5 \right) ( x - 3 )$

$f ( x ) = 4 - ( x + 4 ) ^ { 5 }$ $f ( x ) = 4 - ( x + 4 ) ^ { 5 } A) B) C) D)$

$f ( x ) = - 2 ( x - 5 ) ^ { 5 } + 3$ $f ( x ) = - 2 ( x - 5 ) ^ { 5 } + 3 A) B) C) D)$

Find the x- and y-intercepts of f. - $f ( x ) = 2 x ^ { 4 } ( x - 5 ) ^ { 5 }$

Determine the maximum number of turning points of f. -f(x) = (x - 3) ²(x + 5) ²

Find the x- and y-intercepts of f. - $f ( x ) = - x ^ { 2 } ( x + 6 ) \left( x ^ { 2 } + 1 \right)$ $( x ) = - x ^ { 2 } ( x + 6 ) \left( x ^ { 2 } + 1 \right)$

$f ( x ) = \frac { 1 } { 5 } x ^ { 4 } \left( x ^ { 2 } - 5 \right)$

Determine the maximum number of turning points of f. - $f ( x ) = ( x - 4 ) ^ { 3 }$

Find the x- and y-intercepts of f. - $f ( x ) = ( x + 5 ) ( x - 3 ) ( x + 3 )$

Find the x- and y-intercepts of f. - $f ( x ) = ( x + 1 ) ( x - 2 ) ( x - 1 ) ^ { 2 }$

Find the x- and y-intercepts of f. - $f ( x ) = ( x + 6 ) ^ { 2 }$

Determine the maximum number of turning points of f. - $f ( x ) = ( x - 3 ) ^ { 2 } ( x + 5 ) ^ { 2 }$

Quiz 3: Polynomial and Rational Functions

Determine the maximum number of turning points of f. -f(x) = -x2 (x + 4) 3(x2 - 1)

Find the x- and y-intercepts of f. - f(x) =6x−x3f ( x ) = 6 x - x ^ { 3 }f(x) =6x−x3

f(x) =5(x2+6) (x2+4) 2f ( x ) = 5 \left( x ^ { 2 } + 6 \right) \left( x ^ { 2 } + 4 \right) ^ { 2 }f(x) =5(x2+6) (x2+4) 2

Determine the maximum number of turning points of f. - f(x) =(x+19) 4(x−4) 3f ( x ) = \left( x + \frac { 1 } { 9 } \right) ^ { 4 } ( x - 4 ) ^ { 3 }f(x) =(x+91​) 4(x−4) 3

Find the x- and y-intercepts of f. - f(x) =−x2(x+6) (x2−1) f ( x ) = - x ^ { 2 } ( x + 6 ) \left( x ^ { 2 } - 1 \right) f(x) =−x2(x+6) (x2−1)

Find the x- and y-intercepts of f. -f(x) = (x - 3) (x - 1)

Determine the maximum number of turning points of f. -f(x) = 8x - x3

f(x) =13x2(x2−5) (x−3) f ( x ) = \frac { 1 } { 3 } x ^ { 2 } \left( x ^ { 2 } - 5 \right) ( x - 3 ) f(x) =31​x2(x2−5) (x−3)

f(x) =4−(x+4) 5f ( x ) = 4 - ( x + 4 ) ^ { 5 }f(x) =4−(x+4) 5

f(x) =−2(x−5) 5+3f ( x ) = - 2 ( x - 5 ) ^ { 5 } + 3f(x) =−2(x−5) 5+3

Find the x- and y-intercepts of f. - f(x) =2x4(x−5) 5f ( x ) = 2 x ^ { 4 } ( x - 5 ) ^ { 5 }f(x) =2x4(x−5) 5

Determine the maximum number of turning points of f. -f(x) = (x - 3) 2(x + 5) 2

Find the x- and y-intercepts of f. - f(x) =−x2(x+6) (x2+1) f ( x ) = - x ^ { 2 } ( x + 6 ) \left( x ^ { 2 } + 1 \right) f(x) =−x2(x+6) (x2+1) (x) =−x2(x+6) (x2+1) ( x ) = - x ^ { 2 } ( x + 6 ) \left( x ^ { 2 } + 1 \right) (x) =−x2(x+6) (x2+1)

f(x) =15x4(x2−5) f ( x ) = \frac { 1 } { 5 } x ^ { 4 } \left( x ^ { 2 } - 5 \right) f(x) =51​x4(x2−5)

Determine the maximum number of turning points of f. - f(x) =(x−4) 3f ( x ) = ( x - 4 ) ^ { 3 }f(x) =(x−4) 3

Find the x- and y-intercepts of f. - f(x) =(x+5) (x−3) (x+3) f ( x ) = ( x + 5 ) ( x - 3 ) ( x + 3 ) f(x) =(x+5) (x−3) (x+3)

Find the x- and y-intercepts of f. - f(x) =(x+1) (x−2) (x−1) 2f ( x ) = ( x + 1 ) ( x - 2 ) ( x - 1 ) ^ { 2 }f(x) =(x+1) (x−2) (x−1) 2

Find the x- and y-intercepts of f. - f(x) =(x+6) 2f ( x ) = ( x + 6 ) ^ { 2 }f(x) =(x+6) 2

Determine the maximum number of turning points of f. - f(x) =(x−3) 2(x+5) 2f ( x ) = ( x - 3 ) ^ { 2 } ( x + 5 ) ^ { 2 }f(x) =(x−3) 2(x+5) 2