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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 #3 : Graphing Quadratics & Polynomials

Complete the square to calculate the maximum or minimum point of the given function.

x^2+8x-3

Possible Answers:

(-4,-19)

(-4,19)

(4,-19)

(-19,-4)

(4,19)

Correct answer:

(-4,-19)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

x^2+8x-3

First identify the middle term coefficient.

$\text{Middle Term Coefficient}= 8$

Now divide the middle term coefficient by two.

$\frac{8}{2}=4$

From here write the function with the perfect square.

$x^2+8x-3\Rightarrow (x+4)^2-3+4^2$

When simplified the new function is,

$\\(x+4)^2-3+16\\(x+4)^2+13$

Since the x^2 term is positive, the parabola will be opening up. This means that the function has a minimum value at the vertex. To find the value of the vertex set the inside portion of the binomial equal to zero and solve.

$\\x+4=0 \\x+4-4=0-4 \\x=-4$

From here, substitute the the value into the original function.

x^2+8x-3

$\\y=(-4)^2+8(-4)-3 \\y=16-32-3 \\y=-19$

Therefore the minimum value occurs at the point (-4,-19) .

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Example Question #4 : Graphing Quadratics & Polynomials

Complete the square to calculate the maximum or minimum point of the given function.

x^2+6x+2

Possible Answers:

(-7,-3)

(3,-7)

(-3,7)

(3,7)

(-3,-7)

Correct answer:

(-3,-7)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

x^2+6x+2

First identify the middle term coefficient.

$\text{Middle Term Coefficient}= 6$

Now divide the middle term coefficient by two.

$\frac{6}{2}=3$

From here write the function with the perfect square.

$x^2+6x+2\Rightarrow (x+3)^2+6+3^2$

When simplified the new function is,

$\\(x+3)^2+2+9\\(x+3)^2+11$

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

From here, substitute the the value into the original function.

x^2+6x+2

$\\y=(-3)^2+6(-3)+2 \\y=9-18+2 \\y=-7$

Therefore the minimum value occurs at the point (-3,-7) .

Report an Error

Example Question #51 : Seeing Structure In Expressions

Complete the square to calculate the maximum or minimum point of the given function.

x^2-4x+1

Possible Answers:

(2,3)

(-2,-3)

(2,-3)

(-3,2)

(-2,3)

Correct answer:

(2,-3)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

x^2-4x+1

First identify the middle term coefficient.

$\text{Middle Term Coefficient}= -4$

Now divide the middle term coefficient by two.

$\frac{-4}{2}=-2$

From here write the function with the perfect square.

$x^2-4x+1\Rightarrow (x-2)^2+1+(-2)^2$

When simplified the new function is,

$\\(x-2)^2+1+4\\(x-2)^2+5$

$\\x-2=0 \\x-2+2=0+2\\x=2$

From here, substitute the the value into the original function.

x^2-4x+1

$\\y=(2)^2-4(2)+1 \\y=4-8+1 \\y=-3$

Therefore the minimum value occurs at the point (2,-3) .

Report an Error

Example Question #52 : Seeing Structure In Expressions

Complete the square to calculate the maximum or minimum point of the given function.

x^2-6x-6

Possible Answers:

(-3,15)

(-15,3)

(3,-15)

(-3,-15)

(3,15)

Correct answer:

(3,-15)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

x^2-6x-6

First identify the middle term coefficient.

$\text{Middle Term Coefficient}= -6$

Now divide the middle term coefficient by two.

$\frac{-6}{2}=-3$

From here write the function with the perfect square.

$x^2-6x-6\Rightarrow (x-3)^2-6+(-3)^2$

When simplified the new function is,

$\\(x-3)^2-6+9\\(x-3)^2+3$

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

From here, substitute the the value into the original function.

x^2-6x-6

$\\y=(3)^2-6(3)-6 \\y=9-18-6 \\y=-15$

Therefore the minimum value occurs at the point (3,-15) .

Report an Error

Example Question #7 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Complete the square to calculate the maximum or minimum point of the given function.

-x^2-2x-1

Possible Answers:

(0,1)

(-1,0)

(1,0)

(0,-1)

(0,0)

Correct answer:

(-1,0)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

-x^2-2x-1

First factor out a negative one.

-1(x^2+2x+1)

Now identify the middle term coefficient.

$\text{Middle Term Coefficient}= 2$

Now divide the middle term coefficient by two.

$\frac{2}{2}=1$

From here write the function with the perfect square.

$-(x^2+2x+1)\Rightarrow-((x+1)^2+1+1)$

When simplified the new function is,

$\\-((x+1)^2+2)$

Since the x^2 term is negative, the parabola will be opening down. This means that the function has a maximum value at the vertex. To find the value of the vertex set the inside portion of the binomial equal to zero and solve.

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

From here, substitute the the value into the original function.

-x^2-2x-1

$\\y=-(-1)^2-2(-1)-1 \\y=-1+2-1 \\y=0$

Therefore the maximum value occurs at the point (-1,0) .

Report an Error

Example Question #8 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Complete the square to calculate the maximum or minimum point of the given function.

-x^2-4x-1

Possible Answers:

(3,-2)

(2,-3)

(2,3)

(-2,-3)

(-2,3)

Correct answer:

(-2,3)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

-x^2-4x-1

First factor out a negative one.

-1(x^2+4x+1)

Now identify the middle term coefficient.

$\text{Middle Term Coefficient}= 4$

Now divide the middle term coefficient by two.

$\frac{4}{2}=2$

From here write the function with the perfect square.

$-(x^2+4x+1)\Rightarrow-((x+2)^2+1+2^2)$

When simplified the new function is,

$\\-((x+2)^2+5)$

$\\x+2=0 \\x+2-2=0-2 \\x=-2$

From here, substitute the the value into the original function.

-x^2-4x-1

$\\y=-(-2)^2-4(-2)-1 \\y=-4+8-1 \\y=3$

Therefore the maximum value occurs at the point (-2,3) .

Report an Error

Example Question #9 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Complete the square to calculate the maximum or minimum point of the given function.

-x^2-6x-1

Possible Answers:

(-3,8)

(3,8)

(-3,-8)

(8,-3)

(3,-8)

Correct answer:

(-3,8)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

-x^2-6x-1

First factor out a negative one.

-1(x^2+6x+1)

Now identify the middle term coefficient.

$\text{Middle Term Coefficient}= 6$

Now divide the middle term coefficient by two.

$\frac{6}{2}=3$

From here write the function with the perfect square.

$-(x^2+6x+1)\Rightarrow-((x+3)^2+1+3^2)$

When simplified the new function is,

$\\-((x+3)^2+10)$

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

From here, substitute the the value into the original function.

-x^2-6x-1

$\\y=-(-3)^2-6(-3)-1 \\y=-9+18-1 \\y=8$

Therefore the maximum value occurs at the point (-3,8) .

Report an Error

Example Question #10 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Complete the square to calculate the maximum or minimum point of the given function.

-x^2-4x-5

Possible Answers:

(2,-1)

(-2,-1)

(2,1)

(-2,1)

(-1,-2)

Correct answer:

(-2,-1)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

-x^2-4x-5

First factor out a negative one.

-1(x^2+4x+5)

Now identify the middle term coefficient.

$\text{Middle Term Coefficient}= 4$

Now divide the middle term coefficient by two.

$\frac{4}{2}=2$

From here write the function with the perfect square.

$-(x^2+4x+5)\Rightarrow-((x+2)^2+5+2^2)$

When simplified the new function is,

$\\-((x+2)^2+9)$

$\\x+2=0 \\x+2-2=0-2 \\x=-2$

From here, substitute the the value into the original function.

-x^2-4x-5

$\\y=-(-2)^2-4(-2)-5 \\y=-4+8-5 \\y=-1$

Therefore the maximum value occurs at the point (-2,-1) .

Report an Error

Example Question #22 : Graphical Representation Of Functions

Complete the square to calculate the maximum or minimum point of the given function.

-x^2-6x-5

Possible Answers:

(-3,-4)

(4,-3)

(-3,4)

(3,4)

(3,-4)

Correct answer:

(-3,4)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

-x^2-6x-5

First factor out a negative one.

-1(x^2+6x+5)

Now identify the middle term coefficient.

$\text{Middle Term Coefficient}= 6$

Now divide the middle term coefficient by two.

$\frac{6}{2}=3$

From here write the function with the perfect square.

$-(x^2+6x+5)\Rightarrow-((x+3)^2+5+3^2)$

When simplified the new function is,

$\\-((x+3)^2+14)$

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

From here, substitute the the value into the original function.

-x^2-6x-1

$\\y=-(-3)^2-6(-3)-5 \\y=-9+18-5 \\y=4$

Therefore the maximum value occurs at the point (-3,4) .

Report an Error

Example Question #51 : Seeing Structure In Expressions

Complete the square to calculate the maximum or minimum point of the given function.

f(x)=-x^2-8x-2

Possible Answers:

(-4,14)

(4,-14)

(-4,-14)

(14,-4)

(4,14)

Correct answer:

(-4,14)

Explanation:

Completing the square method uses the concept of perfect squares. Recall that a perfect square is in the form,

$(ax\pm b)^2=(ax\pm b)(ax\pm b)$

where when multiplied out,

a^2x^2+2abx+b^2

the middle term coefficient, when divided by two and squared, results in the coefficient of the last term.

$2ab=\left(\frac{2ab}{2} \right )^2=b^2$

Complete the square for this particular function is as follows.

-x^2-8x-2

First factor out a negative one.

-1(x^2+8x+2)

Now identify the middle term coefficient.

$\text{Middle Term Coefficient}= 8$

Now divide the middle term coefficient by two.

$\frac{8}{2}=4$

From here write the function with the perfect square.

$-(x^2+8x+2)\Rightarrow-((x+4)^2+2+4^2)$

When simplified the new function is,

$\\-((x+4)^2+18)$

$\\x+4=0 \\x+4-4=0-4 \\x=-4$

From here, substitute the the value into the original function.

-x^2-8x-2

$\\y=-(-4)^2-8(-4)-2 \\y=-16+32-2 \\y=14$

Therefore the maximum value occurs at the point (-4,14) .

Report an Error

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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 #3 : Graphing Quadratics & Polynomials

Example Question #4 : Graphing Quadratics & Polynomials

Example Question #51 : Seeing Structure In Expressions

Example Question #52 : Seeing Structure In Expressions

Example Question #7 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Example Question #8 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Example Question #9 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Example Question #10 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Example Question #22 : Graphical Representation Of Functions

Example Question #51 : Seeing Structure In Expressions

All Common Core: High School - Algebra Resources

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