Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #8 : Write A Polynomial Function From Its Zeros

Which of the following equations represents a quadratic equations with zeros at  and , and that passes through point ?

Possible Answers:

Correct answer:

Explanation:

When finding the equation of a quadratic from its zeros, the natural first step is to recreate the factored form of a simple quadratic with those zeros. Putting aside the coefficient, a quadratic with zeros at 3 and -6 would factor to:

 

So you know that a possible quadratic for these zeros would be:

Now you need to determine the coefficient of the quadratic, and that’s where the point (2, -24) comes in. That means that if you plug in , the result of the quadratic will be -24.  So you can set up the equation:

, where  is the coefficient you're solving for. And you know that this is true when  so if you plug in 2, you can solve for :

This means that:

So:

And therefore . When you then distribute that coefficient of 3 across the entire quadratic:

 

So:

 

 

 

Example Question #9 : Write A Polynomial Function From Its Zeros

Which of the following represents a quadratic equation with its zeros at  and ?

Possible Answers:

Correct answer:

Explanation:

The important first step of creating a quadratic equation from its zeros is knowing what a zero really is. A zero of a quadratic (or polynomial) is an x-coordinate at which the y-coordinate is equal to 0.  

We find that algebraically by factoring quadratics into the form , and then setting  equal to  and , because in each of those cases and entire parenthetical term would equal 0, and anything times 0 equals 0.

Here the question gives you a head start: we know that the numbers 4 and 5 can go in the  and  spots, because if so we'll have found our zeros. So we can set up the equation:

This satisfies the requirements of zeros, but now we need to expand this equation using FOIL to turn it into a proper quadratic. That means that our quadratic is:

And when we combine like terms it's:

Example Question #10 : Write A Polynomial Function From Its Zeros

Which of the following equations belongs to a quadratic with zeros at x = 2 and x = 7?

Possible Answers:

Correct answer:

Explanation:

The answer is .

A quadratic with zeros at x = 2 and x = 7 would factor to , where it is important to recognize that a is the coefficient. So while you might be looking to simply expand  to , note that none of the options with a simple  term (and not ) directly equal that simple quadratic when set to 0.

 

However, if you multiply  by 2, you get . That’s a simple addition of 18x to both sides of the equation to get to the correct answer: .

 

 

Example Question #11 : Write A Polynomial Function From Its Zeros

Which of the following equations represents a quadratic equation with zeros at x = 3 and x = -6, and that passes through point (2, -24)?

Possible Answers:

Correct answer:

Explanation:

When finding the equation of a quadratic from its zeros, the natural first step is to recreate the factored form of a simple quadratic with those zeros. Putting aside the coefficient, a quadratic with zeros at 3 and -6 would factor to:

So you know that a possible quadratic for these zeros would be:

Now you need to determine the coefficient of the quadratic, and that’s where the point (2, -24) comes in. That means that if you plug in x = 2, the result of the quadratic will be -24.  So you can set up the equation:

, where a is the coefficient you’re solving for. And you know that this is true when x = 2, so if you plug in x = 2 you can solve for a:

So a = 3.  When you then distribute that coefficient of 3 across the original simple quadratic, you have:

So your quadratic is 

 

 

Example Question #11 : Write A Polynomial Function From Its Zeros

Which of the following equations belongs to a quadratic with zeros at  and ?

Possible Answers:

Correct answer:

Explanation:

A quadratic with zeros at 2 and 7 would factor to , where it is important to recognize that  is the coefficient. So while you might be looking to simply expand   to , note that none of the options with a simple  term (and not ) directly equal that simple quadratic when set to 0.

 

However, if you multiply  by 2, you get . That’s a simple addition of 18x to both sides of the equation to get to the correct answer: .

 

 

Example Question #721 : Intermediate Single Variable Algebra

Which of the following represents a quadratic with zeros at 5 and 2 and that passes through point (1, 16)?

Possible Answers:

 

Correct answer:

Explanation:

When you're finding the equation of a quadratic given its zeros, a good start is to ignore the coefficient (you'll return to it later) and to construct the factored form of the quadratic using the zeros. Here with zeros of 5 and 2, that factored form would be:

That then means you can expand it to:

From here, you can see that the answer choices all maintain the same general relationship as this simple quadratic, but have different coefficients. This is where you can use the point (1, 16) to your advantage. This means that if you plug in  to the quadratic, it will produce an answer of . To solve this algebraically, return to using a coefficient in front of your entire quadratic. You'll then have:

Where  is the coefficient you can solve for knowing that the equation will produce a value of  when you input .  So plug that in and you can solve for the coefficient:

Meaning that:

So you know that .

Then you can multiply that coefficient across your simple quadratic to get the answer:

Becomes:

Example Question #5211 : Algebra Ii

The mean of the following numbers is 13. Solve for x.

18, 15, 10, 8, x

Possible Answers:

12

14

15

13

Correct answer:

14

Explanation:

where n equals the number of events.

Example Question #1 : Mean, Standard Deviation, And Normal Distribution

On 3 exams, Willow has an average score of 95 points. Her lowest score was an 90. If this score is dropped, what is her new average?

Possible Answers:

98

97.5

96

100

Correct answer:

97.5

Explanation:

To answer this question, you must first calculate Willow's total points from the 3 exams. The total points will be equal to 3 times the original average:

95*3=285

Now, if the 90 is removed, his new total will be:

285-90=195

With the score dropped, her new total number of exams is 2. Therefore, the new average is:

Example Question #1 : Mean, Standard Deviation, And Normal Distribution

The ages of the students at Rosa Parks High School are normally distributed with a mean of 15.5 and a standard deviation of 0.6 years.

What is the proportion of students that are older than 16.2 years old?

Possible Answers:

5%

2.5%

1%

0%

Correct answer:

2.5%

Explanation:

This question relates to the 689599.7 rule of normal distribution. We know that 95% of the data are within 2 standard deviations from the mean.

In this case this means that 95% of the students are between

15.520.6 and 15.5+20.5 

151.2 and 15+1.2

13.8 and 16.2

Therefore we have 5% of the students that are outside of this range. Since the normal distribution is symmetric, the proportion of students below 13.8 is the same as the proportion of students above 16.2.

Thus the right answer is one half of 5%, or 2.5%.

2.5% of the students are older than 16.2 years old. 

Example Question #98 : Probability

The ages of the students at Central Universityl are normally distributed with a mean of 23.5 and a standard deviation of 2.5 years.

What is the proportion of students that are younger than 18.5 years old?

Possible Answers:

0%

5%

1%

2.5%

Correct answer:

2.5%

Explanation:

This question relates to the 689599.7 rule of normal distribution. We know that 95% of the data are within 2 standard deviations from the mean.

In this case this means that 95% of the students are between

23.52⋅2.5 and 23.5+2⋅2.5

23.5-5 and 23.5+5

18.5 and 28.5

Therefore we have 5% of the students that are outside of this range. Since the normal distribution is symmetric, the proportion of students below 18.5 is the same as the proportion of students above 28.5.

Thus the right answer is half of 5% or 2.5%.

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