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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #3 : Introduction To Functions

Define a function $f(x) = 3x^{3}+ 4x$ .

Is this function even, odd, or neither?

Possible Answers:

Odd

Even

Neither

Correct answer:

Odd

Explanation:

To identify a function f(x) as even odd, or neither, determine f(-x) by replacing with , then simplifying. If f(-x) = f(x) , the function is even; if f(-x) = - f(x) is odd.

$f(x) = 3x^{3}+ 4x$ ,

$f(-x) = 3(-x)^{3}+ 4(-x)$

By the Power of a Product Property,

$f(-x) = 3(-1)^{3} x^{3}+ 4(-x)$

$f(-x) = 3(-1) x^{3}+ 4(-x)$

$f(-x) = -3x^{3}-4x$

$f(-x) = -3x^{3}-4x = -(3x^{3}+ 4x) = -f(x)$ ,

so f(x) is an odd function

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Example Question #7 : Introduction To Functions

Define a function $f(x) = x^{7}+ x^{6}- x$ .

Is this function even, odd, or neither?

Possible Answers:

Neither

Even

Odd

Correct answer:

Neither

Explanation:

To identify a function f(x) as even odd, or neither, determine f(-x) by replacing with , then simplifying. If f(-x) = f(x) , the function is even; if f(-x) = - f(x) is odd.

$f(x) = x^{7}+ x^{6}- x$

$f(-x) = (-x) ^{7}+ (-x) ^{6}- (-x)$

By the Power of a Product Property,

$f(-x) =(-1)^{7} x ^{7}+(-1)^{6} x ^{6}- (-x)$

$f(-x) =(-1) x ^{7}+1 x ^{6}+x$

$f(-x) =-x ^{7}+ x ^{6}+x$

$f(-x) \ne f(x)$ , so f(x) is not an even function.

$- f(x) = - (x^{7}+ x^{6}- x) = - x^{7}- x^{6} + x$ ,

$f(-x) \ne -f(x)$ , so f(x) is not an odd function.

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Example Question #1 : Functions And Graphs

Define a function $f(x) = \log (x^{2})$ .

Is this function even, odd, or neither?

Possible Answers:

Odd

Neither

Even

Correct answer:

Even

Explanation:

To identify a function f(x) as even, odd, or neither, determine f(-x) by replacing with , then simplifying. If f(-x) = f(x) , the function is even; if f(-x) = - f(x) is odd.

$f(x) = \log (x^{2})$

$f(-x) = \log[ (-x)^{2} ] = \log\left ( x^{2} \right ) = f(x)$

f(-x) = f(x) , so f(x) is an even function.

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Example Question #1 : Functions And Graphs

Define a function $f(x) = 4 ^{x}- \frac{1}{4^{x}}$ .

Is this function even, odd, or neither?

Possible Answers:

Even

Neither

Odd

Correct answer:

Odd

Explanation:

To identify a function f(x) as even, odd, or neither, determine f(-x) by replacing with , then simplifying. If f(-x) = f(x) , the function is even; if f(-x) = - f(x) is odd.

$f(x) = 4 ^{x}- \frac{1}{4^{x}}$

$f(-x) = 4 ^{-x}- \frac{1}{4^{-x}}$

$f(-x) = \frac{1}{4^{x}}- 4 ^{x}$

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

Since f(-x) = - f(x) , f(x) is an odd function.

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Example Question #11 : Introduction To Functions

Function

The above table refers to a function f(x) with domain $\left \{ -3, -2, -1, 0, 1, 2, 3 \right \}$ .

Is this function even, odd, or neither?

Possible Answers:

Odd

Even

Neither

Correct answer:

Odd

Explanation:

A function is odd if and only if, for every in its domain, f(-c) = -f(c) ; it is even if and only if, for every in its domain, f(-c) = f(c) . We can see that

f(-3) = 5 = -(-5) = - f(3)

f(-2) = 7 = -(-7) = - f(2)

f(-1) = 2 = -(-2) = - f(1)

f(-0) = 0 = -0 = -f(0)

It follows that f(x) is an odd function.

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Example Question #12 : Functions And Graphs

Function

The above table refers to a function f(x) with domain $\left \{ -3, -2, -1, 0, 1, 2, 3 \right \}$ .

Is this function even, odd, or neither?

Possible Answers:

Neither

Odd

Even

Correct answer:

Neither

Explanation:

A function is odd if and only if, for every in its domain, f(-c) = -f(c) ; it is even if and only if, for every in its domain, f(-c) = f(c) . We can see that

f(-3) = 9 = -(-9) = - f(3) ;

the function cannot be even. This does allow for the function to be odd. However, if f(x) is odd, then, by definition,

f(-0) = -f(0) , or

f(0) = -f(0)

and f(0) is equal to its own opposite - the only such number is 0, so

f(0) = 0 .

This is not the case - f(0) = 1 - so the function is not odd either.

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Example Question #12 : Introduction To Functions

Function

The above table refers to a function f(x) with domain $\left \{ -3, -2, -1, 0, 1, 2, 3 \right \}$ .

Is this function even, odd, or neither?

Possible Answers:

Neither

Odd

Even

Correct answer:

Even

Explanation:

A function f(x) is odd if and only if, for every in its domain, f(-c) = -f(c) ; it is even if and only if, for every in its domain, f(-c) = f(c) . We can see that

f(-3) = 3 = f(3)

f(-2) =9 = f(2)

f(-1) = -1 = f(1)

Of course,

f(0) = f(-0) .

Therefore, f(x) is even by definition.

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Example Question #531 : Algebra Ii

Odd

Which of the following is true of the relation graphed above?

Possible Answers:

It is an odd function

It is not a function

It is a function, but it is neither even nor odd.

It is an even function

Correct answer:

It is an odd function

Explanation:

The relation graphed is a function, as it passes the vertical line test - no vertical line can pass through it more than once, as is demonstrated below:

Odd

Also, it can be seen to be symmetrical about the origin. Consequently, for each in the domain, f(-c) = -f(c) - the function is odd.

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Example Question #15 : Functions And Graphs

Which is a vertical asymptote of the graph of the function $f(x) = \frac{x- 7}{x-6}$ ?

(a) x = 6

(b) x=7

Possible Answers:

(a) only

Neither (a) nor (b)

(b) only

Both (a) and (b)

Correct answer:

(a) only

Explanation:

The vertical asymptote(s) of the graph of a rational function such as f(x) can be found by evaluating the zeroes of the denominator after the rational expression is reduced. The expression is in simplest form, so set the denominator equal to 0 and solve for :

x-6 = 0

x= 6

The graph of f(x) has the line of the equation x= 6 as its only vertical asymptote.

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Example Question #16 : Functions And Graphs

Which of the following is a vertical asymptote of the graph of the function $f(x) = \frac{x^{2} - 11x+ 30}{x-6}$ ?

(a) x= 5

(b) x= 6

Possible Answers:

Neither (a) nor (b)

Both (a) and (b)

(b) only

(a) only

Correct answer:

Neither (a) nor (b)

Explanation:

The vertical asymptote(s) of the graph of a rational function such as f(x) can be found by evaluating the zeroes of the denominator after the rational expression is reduced.

First, factor the numerator. It is a quadratic trinomial with lead term $x^{2}$ , so look to "reverse-FOIL" it as

$x^{2} - 11x+ 30 = (x+a)(x+b)$

by finding two integers with sum and product 30. By trial and error, these integers can be found to be and , so

$x^{2} - 11x+ 30 = (x-5)(x-6)$

Therefore, f(x) can be rewritten as

$f(x) = \frac{(x-5)(x-6)}{x-6}$ .

Cancelling x -6 , this can be seen to be essentially a polynomial function:

f(x) =x-5 ,

which does not have a vertical asymptote.

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Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #3 : Introduction To Functions

Example Question #7 : Introduction To Functions

Example Question #1 : Functions And Graphs

Example Question #1 : Functions And Graphs

Example Question #11 : Introduction To Functions

Example Question #12 : Functions And Graphs

Example Question #12 : Introduction To Functions

Example Question #531 : Algebra Ii

Example Question #15 : Functions And Graphs

Example Question #16 : Functions And Graphs

All Algebra II Resources

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