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All New SAT Math - Calculator Resources

2 Diagnostic Tests 27 Practice Tests Question of the Day Flashcards Learn by Concept

Want to review New SAT Math - Calculator but don’t feel like sitting for a whole test at the moment? Varsity Tutors has you covered with thousands of different New SAT Math - Calculator flashcards! Our New SAT Math - Calculator flashcards allow you to practice with as few or as many questions as you like. Get some studying in now with our numerous New SAT Math - Calculator flashcards.

Flashcards: Graphing Quadratics & Polynomials

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

x^2+4x+7

$\text{Minimum}=(2,3)$

$\text{Maximum}=(-2,3)$

$\text{Minimum}=(-2,3)$

$\text{Minimum}=(-2,-3)$

$\text{Maximum}=(-2,-3)$

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Flashcards: Graphing Quadratics & Polynomials

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

x^2+4x+7

Your answer:

$\text{Minimum}=(2,3)$

Your answer:

$\text{Maximum}=(-2,3)$

Your answer:

$\text{Minimum}=(-2,-3)$

Your answer:

$\text{Maximum}=(-2,-3)$

Correct answer:

$\text{Minimum}=(-2,3)$

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+7

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. Remember when adding the new squared term, add it to both sides to keep the equation balanced.

$x^2+4x+7=0\Rightarrow (x+2)^2+2^2+7=0+2^2$

When simplified the new function is,

$\\(x+2)^2+4+7=4 \\(x+2)^2+11=4 \\(x+2)^2=-7$

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.

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

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

x^2+4x+7

$\\y=(-2)^2+4(-2)+7 \\y=4-8+7 \\y=3$

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

Flashcards: Graphing Quadratics & Polynomials

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

x^2+2x-4

(-1,5)

(-1,-5)

(-5,-1)

(1,-5)

(1,5)

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Flashcards: Graphing Quadratics & Polynomials

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

x^2+2x-4

Your answer:

(-1,5)

Your answer:

(-5,-1)

Your answer:

(1,-5)

Your answer:

(1,5)

Correct answer:

(-1,-5)

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-4

First 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. Remember when adding the new squared term, add it to both sides to keep the equation balanced.

$x^2+2x-4=0\Rightarrow (x+1)^2+1^2-4=0+1^2$

When simplified the new function is,

$\\(x+1)^2-4+1 \\(x+1)^2-3$

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

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

x^2+2x-4

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

Therefore the minimum value occurs at the point (-1,-5) .

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Flashcards: Graphing Quadratics & Polynomials

Flashcards: Graphing Quadratics & Polynomials

Flashcards: Graphing Quadratics & Polynomials

Flashcards: Graphing Quadratics & Polynomials

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All New SAT Math - Calculator Resources

Flashcards: Graphing Quadratics & Polynomials

Flashcards: Graphing Quadratics & Polynomials

Flashcards: Graphing Quadratics & Polynomials

Flashcards: Graphing Quadratics & Polynomials

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DMCA Complaint

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