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sergey [27]
3 years ago
8

Question 1 of 10

Mathematics
1 answer:
Archy [21]3 years ago
3 0

Answer:

A. y = 7(x + 1)²-3

Step-by-step explanation:

Parabola:

y = 7x^{2} + 14x + 4

y = 7(x^{2} + 2x) + 4

Putting into vertex form, remember that:

(x + a)^{2} = x^{2} + 2ax + a^{2}

In this question:

x^{2} + 2x, to put into this format:

x^{2} + 2x + 1 = (x + 1)^{2}

We add one inside the parenthesis to do this. The parenthesis is multiplied by 7, so for the equivalent, we also have to subtract 7. Then

Vertex form:

y = 7(x^{2} + 2x + 1) + 4 - 7

y = 7(x + 1)^{2} - 3

So the correct answer is:

A. y = 7(x + 1)²-3

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Which statement is true about the graphed function?
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2 years ago
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calculate the annswer when the largest prime number that is a factor of 35 is multiplied by the smallest prime number that is a
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3 years ago
Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?
likoan [24]

\mathrm{Use\:the\:rational\:root\:theorem}

a_0=15,\:\quad a_n=2

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5

=\left(x+3\right)\left(2x^2-11x+5\right)

Factor: 2x^2-11x+5

2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)

=x\left(2x-1\right)-5\left(2x-1\right)

2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)

\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:}

thus zeros of f(x) is

x=-3,\:x=5,\:x=\frac{1}{2}

5 0
2 years ago
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