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Common Core: High School - Algebra : 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 #151 : New Sat Math Calculator

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

$\displaystyle x^2+8x-3$

Possible Answers:

$\displaystyle (-19,-4)$

(4,19) $\displaystyle (4,19)$

(-4,19) $\displaystyle (-4,19)$

$\displaystyle (-4,-19)$

(4,-19) $\displaystyle (4,-19)$

Correct answer:

$\displaystyle (-4,-19)$

Explanation:

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

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

where when multiplied out,

$\displaystyle a^2x^2+2abx+b^2$

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

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

Complete the square for this particular function is as follows.

$\displaystyle x^2+8x-3$

First identify the middle term coefficient.

$\text{Middle Term Coefficient}= 8$ $\displaystyle \text{Middle Term Coefficient}= 8$

Now divide the middle term coefficient by two.

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

From here write the function with the perfect square.

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

When simplified the new function is,

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

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

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

From here, substitute the the $\displaystyle x$ value into the original function.

$\displaystyle x^2+8x-3$

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

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

Report an Error

Example Question #2 : 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.

$\displaystyle x^2+6x+2$

Possible Answers:

(3,7) $\displaystyle (3,7)$

(3,-7) $\displaystyle (3,-7)$

(-7,-3) $\displaystyle (-7,-3)$

(-3,7) $\displaystyle (-3,7)$

(-3,-7) $\displaystyle (-3,-7)$

Correct answer:

(-3,-7) $\displaystyle (-3,-7)$

Explanation:

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

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

where when multiplied out,

$\displaystyle a^2x^2+2abx+b^2$

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

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

Complete the square for this particular function is as follows.

$\displaystyle x^2+6x+2$

First identify the middle term coefficient.

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

Now divide the middle term coefficient by two.

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

From here write the function with the perfect square.

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

When simplified the new function is,

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

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

From here, substitute the the $\displaystyle x$ value into the original function.

$\displaystyle x^2+6x+2$

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

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

Report an Error

Example Question #5 : Graphing Quadratics & Polynomials

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

$\displaystyle x^2-4x+1$

Possible Answers:

(-3,2) $\displaystyle (-3,2)$

(2,-3) $\displaystyle (2,-3)$

(-2,3) $\displaystyle (-2,3)$

(2,3) $\displaystyle (2,3)$

(-2,-3) $\displaystyle (-2,-3)$

Correct answer:

(2,-3) $\displaystyle (2,-3)$

Explanation:

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

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

where when multiplied out,

$\displaystyle a^2x^2+2abx+b^2$

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

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

Complete the square for this particular function is as follows.

$\displaystyle x^2-4x+1$

First identify the middle term coefficient.

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

Now divide the middle term coefficient by two.

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

From here write the function with the perfect square.

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

When simplified the new function is,

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

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

From here, substitute the the $\displaystyle x$ value into the original function.

$\displaystyle x^2-4x+1$

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

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

Report an Error

Example Question #3 : 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.

$\displaystyle x^2-6x-6$

Possible Answers:

(3,-15) $\displaystyle (3,-15)$

(-3,15) $\displaystyle (-3,15)$

(3,15) $\displaystyle (3,15)$

(-15,3) $\displaystyle (-15,3)$

$\displaystyle (-3,-15)$

Correct answer:

(3,-15) $\displaystyle (3,-15)$

Explanation:

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

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

where when multiplied out,

$\displaystyle a^2x^2+2abx+b^2$

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

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

Complete the square for this particular function is as follows.

$\displaystyle x^2-6x-6$

First identify the middle term coefficient.

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

Now divide the middle term coefficient by two.

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

From here write the function with the perfect square.

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

When simplified the new function is,

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

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

From here, substitute the the $\displaystyle x$ value into the original function.

$\displaystyle x^2-6x-6$

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

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

Report an Error

Example Question #7 : Graphing Quadratics & Polynomials

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

$\displaystyle -x^2-2x-1$

Possible Answers:

(0,-1) $\displaystyle (0,-1)$

(-1,0) $\displaystyle (-1,0)$

(1,0) $\displaystyle (1,0)$

(0,1) $\displaystyle (0,1)$

(0,0) $\displaystyle (0,0)$

Correct answer:

(-1,0) $\displaystyle (-1,0)$

Explanation:

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

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

where when multiplied out,

$\displaystyle a^2x^2+2abx+b^2$

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

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

Complete the square for this particular function is as follows.

$\displaystyle -x^2-2x-1$

First factor out a negative one.

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

Now identify the middle term coefficient.

$\text{Middle Term Coefficient}= 2$ $\displaystyle \text{Middle Term Coefficient}= 2$

Now divide the middle term coefficient by two.

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

From here write the function with the perfect square.

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

When simplified the new function is,

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

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

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

From here, substitute the the $\displaystyle x$ value into the original function.

$\displaystyle -x^2-2x-1$

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

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

Report an Error

Example Question #4 : 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.

$\displaystyle -x^2-4x-1$

Possible Answers:

(3,-2) $\displaystyle (3,-2)$

(-2,-3) $\displaystyle (-2,-3)$

(2,-3) $\displaystyle (2,-3)$

(-2,3) $\displaystyle (-2,3)$

(2,3) $\displaystyle (2,3)$

Correct answer:

(-2,3) $\displaystyle (-2,3)$

Explanation:

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

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

where when multiplied out,

$\displaystyle a^2x^2+2abx+b^2$

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

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

Complete the square for this particular function is as follows.

$\displaystyle -x^2-4x-1$

First factor out a negative one.

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

Now identify the middle term coefficient.

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

Now divide the middle term coefficient by two.

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

From here write the function with the perfect square.

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

When simplified the new function is,

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

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

From here, substitute the the $\displaystyle x$ value into the original function.

$\displaystyle -x^2-4x-1$

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

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

Report an Error

Example Question #5 : 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.

$\displaystyle -x^2-6x-1$

Possible Answers:

(3,8) $\displaystyle (3,8)$

(-3,8) $\displaystyle (-3,8)$

(8,-3) $\displaystyle (8,-3)$

(3,-8) $\displaystyle (3,-8)$

(-3,-8) $\displaystyle (-3,-8)$

Correct answer:

(-3,8) $\displaystyle (-3,8)$

Explanation:

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

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

where when multiplied out,

$\displaystyle a^2x^2+2abx+b^2$

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

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

Complete the square for this particular function is as follows.

$\displaystyle -x^2-6x-1$

First factor out a negative one.

$\displaystyle -1(x^2+6x+1)$

Now identify the middle term coefficient.

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

Now divide the middle term coefficient by two.

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

From here write the function with the perfect square.

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

When simplified the new function is,

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

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

From here, substitute the the $\displaystyle x$ value into the original function.

$\displaystyle -x^2-6x-1$

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

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

Report an Error

Example Question #6 : 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.

$\displaystyle -x^2-4x-5$

Possible Answers:

(-2,1) $\displaystyle (-2,1)$

(-2,-1) $\displaystyle (-2,-1)$

(2,-1) $\displaystyle (2,-1)$

(-1,-2) $\displaystyle (-1,-2)$

(2,1) $\displaystyle (2,1)$

Correct answer:

(-2,-1) $\displaystyle (-2,-1)$

Explanation:

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

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

where when multiplied out,

$\displaystyle a^2x^2+2abx+b^2$

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

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

Complete the square for this particular function is as follows.

$\displaystyle -x^2-4x-5$

First factor out a negative one.

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

Now identify the middle term coefficient.

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

Now divide the middle term coefficient by two.

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

From here write the function with the perfect square.

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

When simplified the new function is,

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

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

From here, substitute the the $\displaystyle x$ value into the original function.

$\displaystyle -x^2-4x-5$

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

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

Report an Error

Example Question #22 : Graphical Representation Of Functions

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

$\displaystyle -x^2-6x-5$

Possible Answers:

(4,-3) $\displaystyle (4,-3)$

(-3,4) $\displaystyle (-3,4)$

(3,-4) $\displaystyle (3,-4)$

(3,4) $\displaystyle (3,4)$

(-3,-4) $\displaystyle (-3,-4)$

Correct answer:

(-3,4) $\displaystyle (-3,4)$

Explanation:

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

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

where when multiplied out,

$\displaystyle a^2x^2+2abx+b^2$

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

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

Complete the square for this particular function is as follows.

$\displaystyle -x^2-6x-5$

First factor out a negative one.

$\displaystyle -1(x^2+6x+5)$

Now identify the middle term coefficient.

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

Now divide the middle term coefficient by two.

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

From here write the function with the perfect square.

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

When simplified the new function is,

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

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

From here, substitute the the $\displaystyle x$ value into the original function.

$\displaystyle -x^2-6x-1$

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

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

Report an Error

Example Question #1 : Graphing Quadratics & Polynomials

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

$\displaystyle f(x)=-x^2-8x-2$

Possible Answers:

(14,-4) $\displaystyle (14,-4)$

$\displaystyle (-4,-14)$

(4,14) $\displaystyle (4,14)$

(4,-14) $\displaystyle (4,-14)$

(-4,14) $\displaystyle (-4,14)$

Correct answer:

(-4,14) $\displaystyle (-4,14)$

Explanation:

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

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

where when multiplied out,

$\displaystyle a^2x^2+2abx+b^2$

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

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

Complete the square for this particular function is as follows.

$\displaystyle -x^2-8x-2$

First factor out a negative one.

$\displaystyle -1(x^2+8x+2)$

Now identify the middle term coefficient.

$\text{Middle Term Coefficient}= 8$ $\displaystyle \text{Middle Term Coefficient}= 8$

Now divide the middle term coefficient by two.

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

From here write the function with the perfect square.

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

When simplified the new function is,

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

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

From here, substitute the the $\displaystyle x$ value into the original function.

$\displaystyle -x^2-8x-2$

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

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

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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 #151 : New Sat Math Calculator

Example Question #2 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Example Question #5 : Graphing Quadratics & Polynomials

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

Example Question #7 : Graphing Quadratics & Polynomials

Example Question #4 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Example Question #5 : Calculate Maximum Or Minimum Of Quadratic By Completing The Square: Ccss.Math.Content.Hsa Sse.B.3b

Example Question #6 : 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 #1 : Graphing Quadratics & Polynomials

All Common Core: High School - Algebra Resources

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