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Algebra 1 : Algebra 1

Study concepts, example questions & explanations for Algebra 1

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10 Diagnostic Tests 557 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #77 : Algebraic Functions

Given a function $f(x)=x^{3}-\frac{1}{x^{2}}$ , what is f(2) ?

Possible Answers:

$\frac{33}{4}$

$\frac{30}{4}$

$\frac{32}{4}$

$\frac{29}{4}$

$\frac{31}{4}$

Correct answer:

$\frac{31}{4}$

Explanation:

Given a function

$f(x)=x^{3}-\frac{1}{x^{2}}$ ,

we can plug in x=2 to get

$f(2)=(2)^{3}-\frac{1}{(2)^{2}}=8-\frac{1}{4}=\frac{32}{4}-\frac{1}{4}=\frac{31}{4}$ .

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Example Question #78 : Algebraic Functions

Given a function $f(x)=2x^{2}-3x$ , what is f(10) ?

Possible Answers:

180

170

150

190

160

Correct answer:

170

Explanation:

Given a function

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

we can plug in x=10 to get

$f(10)=2(10)^{2}-3(10)=200-30=170$ .

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Example Question #79 : Algebraic Functions

Given a function $f(x)=x-(\frac{x^{2}-1}{3})$ , what is f(5) ?

Possible Answers:

$\frac{1}{3}$

$-\frac{1}{3}$

$\frac{5}{3}$

Correct answer:

Explanation:

Given a function

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

we can plug in x=5 to get

$f(5)=5-\left(\frac{(5)^{2}-1}{3}\right)=5-\frac{24}{3}=5-8=-3$

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Example Question #80 : Algebraic Functions

Given the function f(x) = 4x+6 , what is $f(\frac{2}{7})$ ?

Possible Answers:

$\frac{48}{7}$

$\frac{24}{7}$

$\frac{12}{7}$

$\frac{50}{7}$

Correct answer:

$\frac{50}{7}$

Explanation:

Start by replacing the number with the term in the function.

$f(\frac{2}{7}) = 4(\frac{2}{7})+6$

Simplify by distributing the four with the two in the numerator.

$4(\frac{2}{7})+6 = \frac{8}{7}+6$

Convert $\frac{6}{1}$ to $\frac{42}{7}$ and replace so that fractions with like denominators can be added.

$\frac{8}{7}+6 = \frac{8}{7}+ \frac{42}{7} = \frac{50}{7}$

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Example Question #81 : Algebraic Functions

f(z)=4z-10

and

$x\diamond y=x^{-y}$ ,

then what is the value of $f(5)\diamond 2$ ?

Possible Answers:

0.1

0.01

100

$\frac{1}{20}$

Correct answer:

0.01

Explanation:

This type of problem is very common on standardized tests, and examines a student's confidence level when dealing with unfamiliar material. Clearly, we don't learn how to "diamond" numbers in class! The diamond symbol, however, just represents some unknown operation. Here, to diamond two numbers simply means raising the first number to the negative second number. The solution is provided below:

First evaluate f(5). This simplifies to:

f(5)=4(5)-10=10 .

Now, we know that the question is asking for the value of:

$10\diamond 2=10^{-2}$ .

The rest of the solution is a simple matter of dealing with a negative exponent. Based on the laws of exponents, we know that

$10^{-2}=\frac{1}{10^2}=\frac{1}{100}=0.01$ .

Therefore, the answer to this question is 0.01.

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Example Question #81 : How To Find F(X)

f(x) = -3x+5

find f(-4) .

Possible Answers:

f(-4) = 17

f(-4) = 12

f(-4) = -4

f(-4) = -7

f(-4) = 8

Correct answer:

f(-4) = 17

Explanation:

Given

f(x) = -3x+5

we must find f(-4) .

All we do is substitute -4 for the variable x.

So anywhere we see x, we will change to -4.

We get

$f(-4) = (-3 \cdot -4) + 5$

f(-4) = 12 + 5

f(-4) = 17

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Example Question #83 : Algebraic Functions

Find f(-2) if

f(x) = 4x^2 - 3x +9 .

Possible Answers:

f(-2) = -1

f(-2) = 31

f(-2) = 19

f(-2) = -13

f(-2) = 23

Correct answer:

f(-2) = 31

Explanation:

Given f(-2) , when solving

f(x) = 4x^2 - 3x +9

we simply substitute.

In this case x = -2 . So anywhere we see an x, we will substitute -2. So,

f(x) = 4x^2 - 3x +9

f(-2) = 4(-2)^2 - 3(-2) + 9

$f(-2) = 4 \cdot 4 + 6 + 9$

f(-2) = 16 + 6 + 9

f(-2) = 31

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Example Question #81 : Algebraic Functions

If x = -4 , solve

f(x) = 2x^2 - 4

Possible Answers:

f(-4) = -28

f(-4) = 60

f(-4) = -36

f(-4) = -68

f(-4) = 28

Correct answer:

f(-4) = 28

Explanation:

When solving a function, we simply substitute the value into the equation. So, we know x = -4 . So in the function

f(x) = 2x^2 - 4

we substitute -4 in for x. By doing that, we get

f(-4) = 2(-4)^2 - 4

We now solve using the order of operations. We get

$f(-4) = 2 \cdot 16 - 4$

f(-4) = 32 - 4

f(-4) = 28

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Example Question #81 : How To Find F(X)

If x = -3 , solve

f(x) = 3x^2 -5x + 1

Possible Answers:

f(-3) = 43

f(-3) = -32

f(-3) = -41

f(-3) = -12

f(-3) = 34

Correct answer:

f(-3) = 43

Explanation:

To solve a function, we simply substitute the value of x into the equation. So, given that

x = -3

we will substitute into the function.

f(x) = 3x^2 -5x + 1

f(-3) = 3(-3)^2-5(-3)+1

f(-3) = 3(9) -5(-3)+1

f(-3) = 27 + 15 + 1

f(-3) = 43

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Example Question #86 : How To Find F(X)

If x = -4 , then solve

f(x) = -2x^2 - 5x + 1

Possible Answers:

f(-4) = -13

f(-4) = 13

f(-4) = -11

f(-4) = -51

f(-4) = 53

Correct answer:

f(-4) = -11

Explanation:

We know x = -4 . When solving a function, we substitute the value of x into the function. In other words, anywhere we see an x, we will replace it with -4.

f(x) = -2x^2 - 5x+1

f(-4) = -2(-4)^2-5(-4)+1

f(-4) = -2(16)+20+1

f(-4) = -32+20+1

f(-4) = -11

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Algebra 1 : Algebra 1

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #77 : Algebraic Functions

Example Question #78 : Algebraic Functions

Example Question #79 : Algebraic Functions

Example Question #80 : Algebraic Functions

Example Question #81 : Algebraic Functions

Example Question #81 : How To Find F(X)

Example Question #83 : Algebraic Functions

Example Question #81 : Algebraic Functions

Example Question #81 : How To Find F(X)

Example Question #86 : How To Find F(X)

All Algebra 1 Resources

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