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Common Core: High School - Algebra : High School: Algebra

Study concepts, example questions & explanations for Common Core: High School - Algebra

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All Common Core: High School - Algebra Resources

8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the -intercept(s) of the function?

y=x^2+2x+1

Possible Answers:

$\text{x-intercept: }-1$

$\text{x-intercept: }1$

$\text{x-intercept: }0$

$\text{x-intercepts: }-1,0,1$

$\text{x-intercepts: }-1,1$

Correct answer:

$\text{x-intercept: }-1$

Explanation:

To find the -intercept of a function, first recall that the -intercept represents the points where the graph of the function crosses the -axis. In other words where the function has a value equal to zero.

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

c_1 and c_2 are factors of and when added together results in .

For the given function,

y=x^2+2x+1

the coefficients are,

$\\b=2 \\c=1$

therefore the factors of that have a sum of are,

y=(x+1)(x+1)

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\\x+1=0\rightarrow x=-1$

To verify, graph the function.

Screen shot 2016 03 08 at 1.07.18 pm

The graph crosses the -axis at -1, thus verifying the result found by factorization.

Report an Error

Example Question #3 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the -intercept(s) of the function?

y=x^2+7x+6

Possible Answers:

$\text{x-intercepts: }-6,1$

$\text{x-intercepts: }-6,-1$

$\text{x-intercepts: }6,-1$

$\text{x-intercepts: }3,2$

$\text{x-intercepts: }6,1$

Correct answer:

$\text{x-intercepts: }-6,-1$

Explanation:

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

c_1 and c_2 are factors of and when added together results in .

For the given function,

y=x^2+7x+6

the coefficients are,

$\\b=7 \\c=6$

therefore the factors of that have a sum of are,

y=(x+1)(x+6)

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\\x+1=0\rightarrow x=-1 \\x+6=0\rightarrow x=-6$

To verify, graph the function.

Screen shot 2016 03 08 at 1.27.02 pm

The graph crosses the -axis at -6 and -1, thus verifying the results found by factorization.

Report an Error

Example Question #221 : High School: Algebra

What are the -intercept(s) of the function?

y=x^2+4x+3

Possible Answers:

$\text{x-intercepts:}-3,-1$

$\text{x-intercepts:}-3,-1,1$

$\text{x-intercepts: }3,-1$

$\text{x-intercepts:}-3,1$

$\text{x-intercepts: }3,1$

Correct answer:

$\text{x-intercepts:}-3,-1$

Explanation:

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

c_1 and c_2 are factors of and when added together results in .

For the given function,

y=x^2+4x+3

the coefficients are,

$\\b=4 \\c=3$

therefore the factors of that have a sum of are,

y=(x+1)(x+3)

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\\x+1=0\rightarrow x=-1 \\x+3=0\rightarrow x=-3$

To verify, graph the function.

Screen shot 2016 03 08 at 1.52.17 pm

The graph crosses the -axis at -1 and -3, thus verifying the results found by factorization.

Report an Error

Example Question #6 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the -intercept(s) of the function?

y=x^2-2x+1

Possible Answers:

$\text{x-intercept: }0$

$\text{x-intercept: }-1$

$\text{x-intercepts: }0,1$

$\text{x-intercepts: }1,-1$

$\text{x-intercept: }1$

Correct answer:

$\text{x-intercept: }1$

Explanation:

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

c_1 and c_2 are factors of and when added together results in .

For the given function,

y=x^2-2x+1

the coefficients are,

$\\b=-2 \\c=1$

therefore the factors of that have a sum of are,

y=(x-1)(x-1)

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\\x-1=0\rightarrow x=1$

To verify, graph the function.

Screen shot 2016 03 09 at 9.54.14 am

The graph crosses the -axis at 1, thus verifying the result found by factorization.

Report an Error

Example Question #7 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the -intercept(s) of the function?

y=x^2-8x+7

Possible Answers:

$\text{x-intercepts: }-1,-7$

$\text{x-intercepts: }1,7$

$\text{x-intercepts: }1,8$

$\text{x-intercepts: }-1,7$

$\text{x-intercepts: }1,-7$

Correct answer:

$\text{x-intercepts: }1,7$

Explanation:

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

c_1 and c_2 are factors of and when added together results in .

For the given function,

y=x^2-8x+7

the coefficients are,

$\\b=-8 \\c=7$

therefore the factors of that have a sum of are,

y=(x-1)(x-7)

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\\x-1=0\rightarrow x=1 \\x-7=0\rightarrow x=7$

To verify, graph the function.

Screen shot 2016 03 09 at 10.02.27 am

The graph crosses the -axis at 1 and 7, thus verifying the results found by factorization.

Report an Error

Example Question #222 : High School: Algebra

What are the -intercept(s) of the function?

y=x^2-5x+6

Possible Answers:

$\text{x-intercepts: }2,1$

$\text{x-intercepts: }-2,-3$

$\text{x-intercepts: }-2,3$

$\text{x-intercepts: }2,-3$

$\text{x-intercepts: }2,3$

Correct answer:

$\text{x-intercepts: }2,3$

Explanation:

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

c_1 and c_2 are factors of and when added together results in .

For the given function,

y=x^2-5x+6

the coefficients are,

$\\b=-5 \\c=6$

therefore the factors of that have a sum of are,

y=(x-2)(x-3)

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\\x-2=0\rightarrow x=2 \\x-3=0\rightarrow x=3$

To verify, graph the function.

Screen shot 2016 03 09 at 10.10.58 am

The graph crosses the -axis at 2 and 3, thus verifying the results found by factorization.

Report an Error

Example Question #11 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

What are the -intercept(s) of the function?

y=x^2+12x+36

Possible Answers:

$\text{x-intercepts: }-6,6$

$\text{x-intercepts: }-6,-1$

$\text{x-intercepts: }-6,1$

$\text{x-intercept: }-6$

$\text{x-intercept: }6$

Correct answer:

$\text{x-intercept: }-6$

Explanation:

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

c_1 and c_2 are factors of and when added together results in .

For the given function,

y=x^2+12x+36

the coefficients are,

$\\b=12 \\c=36$

therefore the factors of that have a sum of are,

y=(x+6)(x+6)

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\\x+6=0\rightarrow x=-6$

To verify, graph the function.

Screen shot 2016 03 09 at 10.18.10 am

The graph crosses the -axis at -6, thus verifying the result found by factorization.

Report an Error

Example Question #222 : High School: Algebra

Find the zeros of $x^{2} + 35 x - 67$

Possible Answers:

$\sqrt{1493} , - \sqrt{1493}$

$\frac{35}{2} + \frac{\sqrt{1493}}{2} , - \frac{\sqrt{1493}}{2} - \frac{35}{2}$

There are no real roots

-35 , 35

Correct answer:

$\frac{35}{2} + \frac{\sqrt{1493}}{2} , - \frac{\sqrt{1493}}{2} - \frac{35}{2}$

Explanation:

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to the coefficients in the equation.

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 35 and c = -67

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -35 \pm \sqrt{ 35 ^2 - 4 * 1 * -67 }}{ 2 * 1 }$
$x =\frac{ -35 \pm \sqrt{ 1225 - -268 }}{ 2 }$
$x =\frac{ -35 \pm \sqrt{ 1493 }}{ 2 }$
$x =\frac{ -35 \pm \sqrt{1493} }{ 2 }$

Now we split this up into two equations.

$x =\frac{ -35 + \sqrt{1493} }{ 2 }$

$x =\frac{ -35 - \sqrt{1493} }{ 2 }$

So our zeros are at

$x = - \frac{35}{2} + \frac{\sqrt{1493}}{2}$

$x = - \frac{\sqrt{1493}}{2} - \frac{35}{2}$

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Example Question #223 : High School: Algebra

Find the zeros of $x^{2} + 35 x - 67$

Possible Answers:

There are no real roots

$\frac{35}{2} + \frac{\sqrt{1493}}{2} , - \frac{\sqrt{1493}}{2} - \frac{35}{2}$

$\sqrt{1493} , - \sqrt{1493}$

Correct answer:

$\frac{35}{2} + \frac{\sqrt{1493}}{2} , - \frac{\sqrt{1493}}{2} - \frac{35}{2}$

Explanation:

$a x^{2} + b x + c = 0$

In this case a = 1 , b = 35 and c = -67

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -35 \pm \sqrt{ 35 ^2 - 4 * 1 * -67 }}{ 2 * 1 }$

$x =\frac{ -35 \pm \sqrt{ 1225 - -268 }}{ 2 }$

$x =\frac{ -35 \pm \sqrt{ 1493 }}{ 2 }$

$x =\frac{ -35 \pm \sqrt{1493} }{ 2 }$

Now we split this up into two equations.

$x =\frac{ -35 + \sqrt{1493} }{ 2 }$

$x =\frac{ -35 - \sqrt{1493} }{ 2 }$

So our zeros are at

$x = - \frac{35}{2} + \frac{\sqrt{1493}}{2}$

$x = - \frac{\sqrt{1493}}{2} - \frac{35}{2}$

Report an Error

Example Question #11 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

Find the zeros of $- 7 x^{2} + 14 x + 33$

Possible Answers:

$1 + \frac{2 \sqrt{70}}{7} , - \frac{2 \sqrt{70}}{7} + 1$

-14 , 14

$4 \sqrt{70} , - 4 \sqrt{70}$

There are no real roots

Correct answer:

$1 + \frac{2 \sqrt{70}}{7} , - \frac{2 \sqrt{70}}{7} + 1$

Explanation:

$a x^{2} + b x + c = 0$

In this case a = -7 , b = 14 and c = 33

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -14 \pm \sqrt{ 14 ^2 - 4 * -7 * 33 }}{ 2 * -7 }$

$x =\frac{ -14 \pm \sqrt{ 196 - -924 }}{ -14 }$

$x =\frac{ -14 \pm \sqrt{ 1120 }}{ -14 }$

$x =\frac{ -14 \pm 4 \sqrt{70} }{ -14 }$

Now we split this up into two equations.

$x =\frac{ -14 + 4 \sqrt{70} }{ -14 }$

$x =\frac{ -14 - 4 \sqrt{70} }{ -14 }$

So our zeros are at

$x = 1 + \frac{2 \sqrt{70}}{7}$

$x = - \frac{2 \sqrt{70}}{7} + 1$

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : High School: Algebra

Study concepts, example questions & explanations for Common Core: High School - Algebra

All Common Core: High School - Algebra Resources

Example Questions

Example Question #2 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

Example Question #3 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

Example Question #221 : High School: Algebra

Example Question #6 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

Example Question #7 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

Example Question #222 : High School: Algebra

Example Question #11 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

Example Question #222 : High School: Algebra

Example Question #223 : High School: Algebra

Example Question #11 : Identify Zeros, Factor And Graph Polynomials: Ccss.Math.Content.Hsa Apr.B.3

All Common Core: High School - Algebra Resources

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