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Which equation is
y=9x^(2)+9x-1 rewritten in vertex form?

y=9(x+(1)/(2))^(2)-(13)/(4)

y=9(x+(1)/(2))^(2)-1

y=9(x+(1)/(2))^(2)+(5)/(4)

y=9(x+(1)/(2))^(2)-(5)/(4)

Which equation is $y=9 x^{2}+9 x-1$ rewritten in vertex form? $\newline$ $y=9\left(x+\frac{1}{2}\right)^{2}-\frac{13}{4}$ $\newline$ $y=9\left(x+\frac{1}{2}\right)^{2}-1$ $\newline$ $y=9\left(x+\frac{1}{2}\right)^{2}+\frac{5}{4}$ $\newline$ $y=9\left(x+\frac{1}{2}\right)^{2}-\frac{5}{4}$

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Convert equations of parabolas from general to vertex form

Full solution

Q. Which equation is

y=9 x^{2}+9 x-1

rewritten in vertex form?

\newline

y=9\left(x+\frac{1}{2}\right)^{2}-\frac{13}{4}

\newline

y=9\left(x+\frac{1}{2}\right)^{2}-1

\newline

y=9\left(x+\frac{1}{2}\right)^{2}+\frac{5}{4}

\newline

y=9\left(x+\frac{1}{2}\right)^{2}-\frac{5}{4}

Identify vertex form: Identify the vertex form of a parabola, which is $y = a(x - h)^2 + k$ .
Start with given equation: Start with the given equation $y = 9x^2 + 9x - 1$ .
Factor out coefficient: Factor out the coefficient of $x^2$ from the first two terms: $y = 9(x^2 + x) - 1$ .
Find completing square value: Find the value to complete the square: $(\frac{b}{2a})^2 = (\frac{1}{2})^2 = \frac{1}{4}$ .
Add and subtract values: Add and subtract $(1/2)^2$ inside the parentheses: $y = 9(x^2 + x + 1/4 - 1/4) - 1$ .
Simplify inside parentheses: Simplify inside the parentheses: $y = 9((x + \frac{1}{2})^2 - \frac{1}{4}) - 1$ .
Distribute the $9$ : Distribute the $9$ : $y = 9(x + \frac{1}{2})^2 - 9(\frac{1}{4}) - 1$ .
Combine the constants: Combine the constants: $-9(\frac{1}{4}) - 1 = -\frac{9}{4} - \frac{4}{4} = -\frac{13}{4}$ .
Write in vertex form: Write the equation in vertex form: $y = 9(x + \frac{1}{2})^2 - \frac{13}{4}$ .

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