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College Algebra : College Algebra

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Example Questions

Example Question #171 : College Algebra

Function

The above diagram shows the graph of function f(x) on the coordinate axes. True or false: The -intercept of the graph is (0, 3)

Possible Answers:

False

True

Correct answer:

False

Explanation:

The -intercept of the graph of a function is the point at which it intersects the -axis (the vertical axis). That point is marked on the diagram below:

Function

The point is about one and three-fourths units above the origin, making the coordinates of the -intercept $\left ( 0, 1\frac{3}{4}\right )$ .

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Example Question #4 : Miscellaneous Functions

Function

A function f(x) is defined on the domain $\left \{ 1, 2, 3, 4, 5\right \}$ according to the above table.

Define a function g(x) = 3x- 5 . Which of the following values is not in the range of the function $(g \circ f )(x)$ ?

Possible Answers:

Correct answer:

Explanation:

This is the composition of two functions. By definition, $(g \circ f )(x) = g \left [ f(x)\right ]$ . To find the range of $(g \circ f )(x)$ , we need to find the values of this function for each value in the domain of f(x) . Since $(g \circ f )(x) = g \left [ f(x)\right ]$ , this is equivalent to evaluating g(x) for each value in the range of f(x) , as follows:

g(x) = 3x- 5

Range value: 3

g(3) = 3 (3)- 5 = 9 - 5 = 4

Range value: 5

g(5) = 3 (5)- 5 = 15 - 5 = 10

Range value: 8

g(8) = 3 (8)- 5 = 24 - 5 = 19

Range value: 13

g(13) = 3 (13)- 5 = 39 - 5 = 34

Range value: 21

g(21) = 3 (21)- 5 = 63 - 5 = 58

The range of g(x) on the set of range values of f(x) - and consequently, the range of $\left (g \circ f \right ) (x)$ - is the set $\left \{4 , 10, 19, 34, 58\right \}$ . Of the five choices, only 45 does not appear in this set; this is the correct choice.

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Example Question #91 : Graphs

Evaluate: $\sum_{k =2}^{5} (-1)^{k}(3k+1)$

Possible Answers:

Correct answer:

Explanation:

Evaluate the expression $(-1)^{k}(3k+1)$ for k = 2, 3, 4, 5 , then add the four numbers:

$a_{k} = (-1)^{k}(3k+1)$

$a_{2} = (-1)^{2}(3 \cdot 2 +1) = 1 (6+ 1) = 7$

$a_{3} = (-1)^{3}(3 \cdot 3 +1) = -1 (9+ 1) = -10$

$a_{4} = (-1)^{4}(3 \cdot 4 +1) = 1 (12+ 1) = 13$

$a_{5} = (-1)^{5}(3 \cdot 5 +1) =- 1 (15+ 1) = -16$

$\sum_{k =2}^{5} (-1)^{k}(3k+1) = 7 +( -10) + 13 + (-16 ) = -6$

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Example Question #2 : Miscellaneous Functions

Evaluate: $\sum_{k =3}^{7} (-2)^{k}$

Possible Answers:

248

Correct answer:

Explanation:

Evaluate the expression $\sum_{k =3}^{7} (-2)^{k}$ for k = 3, 4, 5, 6, 7 , then add the five numbers:

$a_{n} = (-2)^{k}$

$a_{3} = (-2)^{3} = -8$

$a_{4} = (-2)^{4 } =16$

$a_{5} = (-2)^{5 } =-32$

$a_{6} = (-2)^{6 } =64$

$a_{7} = (-2)^{7 } =-128$

$\sum_{k =3}^{7} (-2)^{k} = -8 + 16 + (-32)+ 64 +( -128 ) = -88$

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Example Question #2 : Miscellaneous Functions

$\left \lfloor N \right \rfloor$ refers to the floor of , the greatest integer less than or equal to .

$\left \lceil N \right \rceil$ refers to the ceiling of , the least integer greater than or equal to .

Define $f(x) = \left \lfloor \frac{1}{2}x- \frac{2}{3} \right \rfloor$ and $g(x) = \left \lceil \frac{2}{3}x- \frac{1}{2} \right \rceil$

Which of the following is equal to $(f \circ g ) \left ( \frac{11}{4} \right )$ ?

Possible Answers:

Correct answer:

Explanation:

$(f \circ g ) \left ( \frac{11}{4} \right ) =f\left ( g \left ( \frac{11}{4} \right ) \right )$ , so, first, evaluate $g \left ( \frac{11}{4} \right )$ by substitution:

$g(x) = \left \lceil \frac{2}{3}x- \frac{1}{2} \right \rceil$

$g \left ( \frac{11}{4} \right ) = \left \lceil \frac{2}{3} \left ( \frac{11}{4} \right ) - \frac{1}{2} \right \rceil$

$= \left \lceil \frac{11}{6} - \frac{1}{2} \right \rceil$

$= \left \lceil \frac{4}{3} \right \rceil$

= 2

$(f \circ g ) \left ( \frac{11}{4} \right ) =f\left ( g \left ( \frac{11}{4} \right ) \right ) = f(2)$ , so evaluate f(2) by substitution.

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

$f(2) = \left \lfloor \frac{1}{2} (2)- \frac{2}{3} \right \rfloor$

$= \left \lfloor1 - \frac{2}{3} \right \rfloor$

$= \left \lfloor \frac{1}{3} \right \rfloor$

= 0 ,

the correct response.

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Example Question #3 : Miscellaneous Functions

$\left \lfloor N \right \rfloor$ refers to the floor of , the greatest integer less than or equal to .

$\left \lceil N \right \rceil$ refers to the ceiling of , the least integer greater than or equal to .

Define $f(x) = \left \lceil 2.2x +6.3 \right \rceil$ and $g (x) = \left \lfloor -2.1 x + 9.3 \right \rfloor$ .

Evaluate $(f \circ g )(3.4 )$

Possible Answers:

Correct answer:

Explanation:

$(f \circ g )(3.4 ) = f(g(3.4))$ , so first, evaluate g(3.4 ) using substitution:

$g (x) = \left \lfloor -2.1 x + 9.3 \right \rfloor$

$g (3.4) = \left \lfloor -2.1 (3.4) + 9.3 \right \rfloor$

$= \left \lfloor -7.14+ 9.3 \right \rfloor$

$= \left \lfloor 2.16 \right \rfloor$

= 2

$(f \circ g )(3.4 ) = f(g(3.4)) = f(2)$ , so evaluate f(2) using substitution:

$f(x) = \left \lceil 2.2x +6.3 \right \rceil$

$= \left \lceil 2.2 (2) +6.3 \right \rceil$

$= \left \lceil 4.4 +6.3 \right \rceil$

$= \left \lceil 10.7 \right \rceil$

= 11 ,

the correct response.

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Example Question #172 : College Algebra

Consider the polynomial

$2x^{5} + 3 x^{2}- 4x + k$ ,

where is a real constant. For x- 2 to be a zero of this polynomial, what must be?

Possible Answers:

None of the other choices gives the correct response.

k= -44

k= 68

k= 44

k = -68

Correct answer:

k = -68

Explanation:

By the Factor Theorem, x- c is a zero of a polynomial P(x) if and only if P(c) = 0 . Here, c = 2 , so evaluate the polynomial, in terms of , for x=2 by substituting 2 for :

$2 \cdot 2^{5} + 3 \cdot 2^{2}- 4 \cdot 2 + k$

$= 2 \cdot 32 + 3 \cdot 4 - 4 \cdot 2 + k$

=64 + 12 - 8 + k

=68 + k

Set this equal to 0:

68 + k = 0

k = -68

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Example Question #1 : Simplifying Polynomials

Divide the trinomial below by .

$18x^{2}+ 9x -3$

Possible Answers:

-x+1

$2x+1-\frac{1}{3x}$

$2x+1-\frac{3}{x}$

$2x^{2}+ x -\frac{1}{3}$

$-3x+1+\frac{1}{2x}$

Correct answer:

$2x+1-\frac{1}{3x}$

Explanation:

$18x^{2}+ 9x -3$

We can accomplish this division by re-writing the problem as a fraction.

$\frac{18x^2+9x-3}{9x}$

The denominator will distribute, allowing us to address each element separately.

$\frac{18x^{2}+ 9x -3}{9x}=\frac{18x^{2}}{9x}+\frac{ 9x}{9x}-\frac{3}{9x}$

Now we can cancel common factors to find our answer.

$(\frac{18}{9}* \frac{x^{2}}{x})+1-(\frac{3}{9}* \frac{1}{x})$

$2x+1-\frac{1}{3x}$

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Example Question #172 : College Algebra

Simplify:

$y=\frac{(x^2-16)(x+7)}{x+4}+(x-1)^2+13$

Possible Answers:

$y=\frac{x^3+7x^2-16x-112}{x+4}+(x-1)^2+13$

y=2(x^2+x-7)

y=2x^2+x-14

y=x^2-x-14

$y=\frac{(x^2-16)(x+7)}{x+4}+x^2-2x+14$

Correct answer:

y=2x^2+x-14

Explanation:

$y=\frac{(x^2-16)(x+7)}{x+4}+(x-1)^2+13$

First, factor the numerator of the quotient term by recognizing the difference of squares:

$y=\frac{(x-4)(x+4)(x+7)}{x+4}+(x-1)^2+13$

Cancel out the common term from the numerator and denominator:

y=(x-4)(x+7)+(x-1)^2+13

FOIL (First Outer Inner Last) the first two terms of the equation:

y=x^2+7x-4x-28+x^2-2x+1+13

Combine like terms:

y=2x^2+x-14

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Example Question #4314 : Algebra 1

Divide:

$(x^{3} -23x+10) \div (x-5)$

Possible Answers:

$x^{2} + 5x + 2 + \frac{20}{x-5}$

$x^{2} + 5x + 2$

$x^{2}+1+\frac{12}{x-5}$

$x^{2} - 5x + 16 + \frac{87}{x-5}$

$x^{2} - 5x + 16 - \frac{73}{x-5}$

Correct answer:

$x^{2} + 5x + 2 + \frac{20}{x-5}$

Explanation:

First, rewrite this problem so that the missing $x^{2}$ term is replaced by $0x^{2}$

$(x^{3} + 0x^{2} -23x+10) \div (x-5)$

Divide the leading coefficients:

$x^{3} \div x = x^{2}$ , the first term of the quotient

Multiply this term by the divisor, and subtract the product from the dividend:

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

$(x^{3} + 0x^{2} -23x+10) - (x^{3} -5x^{2}) = 5x^{2}-23x+10$

Repeat this process with each difference:

$5x^{2} \div x = 5x$ , the second term of the quotient

$5x (x-5) = 5 x^{2} -25$

$5x^{2}-23x+10 - ( 5x^{2} -25x) = 2x+10$

One more time:

$2x \div x = 2$ , the third term of the quotient

2 (x-5) = 2x-10

2x+10 - (2x-10) = 20 , the remainder

The quotient is $x^{2} + 5x + 2$ and the remainder is ; this can be rewritten as a quotient of

$x^{2} + 5x + 2 + \frac{20}{x-5}$

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College Algebra : College Algebra

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #171 : College Algebra

Example Question #4 : Miscellaneous Functions

Example Question #91 : Graphs

Example Question #2 : Miscellaneous Functions

Example Question #2 : Miscellaneous Functions

Example Question #3 : Miscellaneous Functions

Example Question #172 : College Algebra

Example Question #1 : Simplifying Polynomials

Example Question #172 : College Algebra

Example Question #4314 : Algebra 1

All College Algebra Resources

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